why the bodies of water important for recreation​

Answers

Answer 1

Explanation:

Recreational water activities can have substantial benefits to health and well-being. Swimming pools, beaches, lakes, rivers and spas provide environments for rest and relaxation, physical activity, exercise, pleasure and fun. Yet they also present risks to health.

pls make it the brainliest of it has helped you !!!!

Answer 2
Because the ice caps are running out of water

Related Questions

A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational field strength of Planet X?


A. The gravitational field strength of Planet X is mg.

B. The gravitational field strength of Planet X is Wx/m.

C. The gravitational field strength of Planet X is 9.8 N/kg.

D. The gravitational field strength of Planet X is mWx.

Answers

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

A 50 kg bicyclist starts his ride down the road with an acceleration of 1m/s2 in air with a density of 1.2 kg/m3. If his velocity at a given moment is 2m/s, how much force is he exerting? Assume the area of his body is 0.5m^2.
a. The bicyclist is exerting 1.1 N of force.
b. The bicyclist is exerting 49 N of force.
c. The bicyclist is exerting 50 N of force.
d. The bicyclist is exerting 51 N of force.

Answers

Answer:

b. The bicyclist is exerting 49 N of force

Explanation:

Given;

mass of bicyclist start, m = 50 kg

acceleration, a = 1 m/s²

density of air, ρ = 1.2 kg/m³

velocity, v = 2 m/s

Area of the bicyclist body, A = 0.5 m²

The drag force on the bicyclist is given by ;

Fd = 0.5CρAv²

where;

C is drag coefficient = 0.9 for bicycle

Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²

Fd = 1.1 N

The force of the bicyclist is given by;

F = ma

F = 50 x 1

F = 50 N

The effective force exerted by the bicyclist is given by;

Fe = F - Fd = 50 N - 1.1 N

Fe = 49 N

Therefore, the force exerted by the bicyclist is 49 N

How can you tell if hair was forcibly removed?

Answers

Answer:Explanation:

A microscopic hair examination can also determine if a hair was forcibly removed, artificially treated or diseased. A comparison microscope can be used to compare a questioned hair to a known hair sample in order to determine if the hairs are similar and if they could have come from a common source.

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth’s electric field a feasible means of flight? Why or why not?

Answers

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

[tex]\Sigma F = F_{E}-W = 0[/tex] (Eq. 1)

Where:

[tex]F_{E}[/tex] - Electrostatic force exerted on human, measured in Newton.

[tex]W[/tex] - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

[tex]q\cdot E-m\cdot g = 0[/tex]

[tex]q = \frac{m\cdot g}{E}[/tex] (Eq. 2)

[tex]E[/tex] - Electric field, measured in Newtons per Coloumb.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravity acceleration, measured in meters per square second.

[tex]q[/tex] - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that [tex]m = 60\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]E = -150\,\frac{N}{C}[/tex], the charge that a 60-kg human must have to overcome weight is:

[tex]q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }[/tex]

[tex]q = -3.923\,C[/tex]

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

[tex]F = \kappa \cdot \frac{q^{2}}{r^{2}}[/tex] (Eq. 3)

Where:

[tex]\kappa[/tex] - Electrostatic constant, measured in Newton-square meter per square Coulomb.

[tex]q[/tex] - Electric charge, measured in Coulomb.

[tex]r[/tex] - Distance between two people, measured in meters.

If we know that [tex]\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}[/tex], [tex]q = -3.923\,C[/tex] and [tex]r = 100\,m[/tex], then the force of repulsion between two people is:

[tex]F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right][/tex]

[tex]F = 13.851\times 10^{6}\,N[/tex]

The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

A 126 N force is applied at an angle of 25.00 to a 8.50 kg block pressed against a rough vertical wall and the block slides down the wall at constant velocity. Calculate the coefficient of kinetic friction between the block and the wall.

Answers

Answer:

μk = 0.58

Explanation:

If the block is sliding down at constant speed, this means that no net force is acting upon it in the vertical direction.As the block is pressed on the wall, this means that it doesnt accelerate in the horizontal direction either, so no net force acts upon it  in this direction also.In this direction, we have only two forces acting, equal and opposite each other, one is the normal force (exerted by the wall) and the other is the horizontal component of the applied force.If the applied force forms an angle of 25º with the wall (which is vertical), this means that we can get its projection along the horizontal direction, using simple trigonometry , as follows:

       [tex]F_{apph} = F_{app} * sin\theta = 126 N * sin 25 = 53.3 N[/tex]

       ⇒  [tex]F_{n} = - F_{apph} = -53.3 N[/tex]

In the vertical direction, we have three forces acting on the block: the weight pointing downward, the kinetic friction force (as we know that the block is sliding), and the vertical component of the applied force, in the same direction as the friction one.As we have already said, the sum of these forces must be 0.[tex]F_{g} + F_{appv} + F_{ff} = 0 (1)[/tex] where Fg is the weight of the block, Fappv is the vertical component of  the applied force, and Fff is the kinetic friction force.Replacing these forces by their mathematical expressions, we have:

       [tex]F_{g} = m_{b} * g = 8.5 Kg * (-9.8 m/s2) = -83.3 N[/tex]

       [tex]F_{appv} = F_{app}* cos\theta = 126 N * cos 25 = 114.2 N[/tex]

      [tex]F_{ff} = \mu_{k}* F_{n} =\mu_{k} * (-53.3 N)[/tex]

Replacing  in (1), and solving for μk, we finally get:

        μk = 0.58

Radio station KBOB broadcasts at a frequency of 85.7 MHz on your dial using radio waves that travel at 3.00 × 108 m/s. Since most of the station's audience is due south of the transmitter, the managers of KBOB don't want to waste any energy broadcasting to the east and west. They decide to build two towers, transmitting in phase at exactly the same frequency, aligned on an east-west axis. For engineering reasons, the two towers must be AT LEAST 10.0 m apart. What is the shortest distance between the towers that will eliminate all broadcast power to the east and west?

Answers

Answer:

12.5 m

Explanation:

The first thing we would do is to calculate the wavelength. To do this, we use the formula

v = fλ, where

v = wave speed

f = frequency

λ = wavelength

If we make wavelength the formula, we have

wavelength = speed / frequency

Now, we substitute the values we had been given and we have

wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m

half of this said wavelength will be

= 3.50 / 2

= 1.75 m

As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,

(10 / 1.75) = 5.7

So the separation will have to be 7 half wavelengths

= (7 * 1.75) = 12.5 m

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *

A. 0.2 m/s
B. 2 m/s
C. 4 m/s
D. 10 m/s

Answers

Answer:

D. 10 m/s

Explanation:

At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?

Answers

Answer:

At a distance of 100 m from the source the intensity will be 40 dB.

Explanation:

Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.

The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.

The conversion between intensity and decibels corresponds to:

[tex]L=10*log\frac{I}{I0}[/tex]

where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.

In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:

[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]

[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]

Being L1= 70 dB and L2= 40 dB

[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]

[tex]30=10*[log(\frac{I1}{I2})][/tex]

[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]

[tex]3=log(\frac{I1}{I2})[/tex]

[tex]10^{3} =\frac{I1}{I2}[/tex]

[tex]1,000=\frac{I1}{I2}[/tex]

The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:

[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]

Then:

[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]

Being d1= 10 m

[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]

[tex]1,000=\frac{d2^{2} }{100}[/tex]

1,000*100= d2²

10,000= d2²

√10,000= d2

100 m= d2

At a distance of 100 m from the source the intensity will be 40 dB.

1 (4 points) A 2-kg ball is moving with a constant speed of 5 m/s in a horizontal circle whose radius is 50 cm. What is the magnitude of the net force on the ball

Answers

Answer:

100 N

Explanation:

The following data were obtained from the question:

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 50 cm

Force (F) =.?

Next, we shall convert 50 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

50 cm = 50 cm × 1 m / 100 cm

50 cm = 0.5 m

Therefore, 50 cm is equivalent to 0.5 m.

Finally, we shall determine the magnitude of the net force on the ball by using the following formula:

F = mv²/r

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 0.5 m

Force (F) =.?

F = mv²/r

F = 2 × 5²/ 0.5

F = 2 × 25/ 0.5

F = 50 / 0.5

F = 100 N

Therefore, the magnitude of the net force on the ball is 100 N.

The magnitude of the net force on the ball will be "100 N".

Force and speed

According to the question,

Mass, m = 2 kg

Velocity, v = 5 m/s

Radius, r = 50 cm or,

               = 50 × [tex]\frac{1}{100}[/tex]

               = 0.5 m

We know the relation,

Force, F = [tex]\frac{mv^2}{r}[/tex]

By substituting the values, we get

              = [tex]\frac{2\times (25)^2}{0.5}[/tex]

              = [tex]\frac{2\times 25}{0.5}[/tex]

              = [tex]\frac{50}{0.5}[/tex]

              = 100 N

Thus the above response is appropriate.

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What is the car’s average velocity (in m/s) in interval between t=1.0s to t=1.5s?

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula for velocity is;

Velocity (in m/s) = distance/time

The distance the car covered in the completed question is divided by the difference in the time interval

The difference in the time interval will be = 1.5s - 1.0s = 0.5s

NOTE: the distance must be in meters or be converted to meters

for an emitted wavelength of 500 nanometers and a redshift of 0.4 what will be the observed wavelength g

Answers

Answer:

The observed wavelength is  [tex] \lambda = 700nm[/tex] (color -  Red)

Explanation:

From the question we are told that

   The wavelength of the emitter is  [tex]\lambda_ e  = 500 nm  =  500 *10^{-9} \  m[/tex]

  The redshift is  R  =  0.4  

     Generally red shift is mathematically represented as

    [tex]R =  \frac{ \lambda -  \lambda_e }{\lambda_e}[/tex]

=>  [tex]0.4  =  \frac{ \lambda -   500 *10^{-9} }{500 *10^{-9} }[/tex]

=>  [tex] \lambda - 500*10^{-9} = 200*10^{-9}  [/tex]

=>   [tex] \lambda = 700 *10^{-9}[/tex]

=>   [tex] \lambda = 700nm[/tex]

A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory with time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?

Answers

Answer:

Answer and steps in the pic

a. In one short sentence, explain why we call the force of gravity an attractive force.
b. Does a force of gravity exist between any two objects

Answers

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G[tex]\frac{m1m2}{r}[/tex]

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

Hollywood and video games often depict the bad guys being "blown away" when they’re shot by a bullet (i.e. once hit, their feet leave the ground and they fly backwards). Assuming that even if a handgun cartridge did generate enough momentum for the bullet to do this, why is it still nonsense on-screen?

Answers

Answer:

Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".

Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.  

What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.  

The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.  

It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this,  it is still nonsense on screen in Hollywood and video.  

             

Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:

τs = -κ θ.

Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.

Required:
a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.
b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device?

Answers

Complete Question

Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:

τs = -κ θ.

Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.

Required:

a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.

b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device? Assume the full scale deflection is 60° from the spring's equilibrium position

Answer:

a

[tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

b

[tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

Explanation:

From the question we are told that

   The number of turns is  N  =  50  

   The radius is  r =  1.55 cm  =  0.0155 m

   The  magnetic field is  B  =  0.35 T

   The induced current is  [tex]I =  45 \mu A =  45 *10^{-6} \  A[/tex]

   

Generally the area of  loop is mathematically represented as

      [tex]A =  \pi r^2[/tex]

=>   [tex]A =3.142 *  0.0155^2[/tex]

=>   [tex]A =0.000755\ m^2[/tex]

Generally the maximum torque is mathematically represented as

     [tex]\tau_{m} =  N  *  B  * I  *  A[/tex]

=>  [tex]\tau_{m} =  50  *  0.35  * 45 *10^{-6} *  0.000755[/tex]

=>  [tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

Generally the  torque 60° from the spring's equilibrium position is mathematically represented as

      [tex]\tau  =  N  *  B  *  I  *  A  *  sin (60)[/tex]

=>    [tex]\tau  =  50  *  0.35  *  45 *10^{-6} *  0.000755  *  sin (60)[/tex]

=>   [tex]\tau  = 2.973 *10^{-7} \  N  \cdot m [/tex]

Generally the toque constant of the spring is mathematically represented as

     [tex]\beta =  \frac{\tau}{60}[/tex]

=>   [tex]\beta =  \frac{\tau}{\frac{\pi}{3}}[/tex]

=>   [tex]\beta =  \frac{2.973 *10^{-7}}{\frac{\pi}{3}}[/tex]

=>    [tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?

Answers

Answer:

The displacement of the canoe is 1.46 m

Explanation:

Given that,

Mass of canoe = 61 kg

Mass of man = 86.5 kg

Length = 4 m

Let the the displacement of the canoe is x'

We need to calculate the displacement of the man

Using formula of displacement

[tex]x=x_{2}-x_{1}[/tex]

Put the value into the formula

[tex]x=4-(0.75+0.75)[/tex]

[tex]x=2.5\ m[/tex]

We need to calculate the displacement of the canoe

Using conservation of momentum

[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]

[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]

[tex]86.5\times2.5=(61+86.5)\times x'[/tex]

[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]

[tex]x'=1.46\ m[/tex]

Hence, The displacement of the canoe is 1.46 m

What are the standard international (si) units of distance

Answers

Answer:

meter

Explanation:

Answer: The International System of Units is a system of measurement based on 7 base units

Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.

HELP PLS GIVING BRAINLIEST!!

Answers

I believe it would be 9ms2 because 10 plus 4 is 14 and 5 plus 4 is 9.

I think it’s 9ms2. Sorry if I’m wrong.

use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is

Answers

Complete question:

use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is 2.2 x 10⁻⁵ km/s/Lyr

Answer:

The distance to the quasar is 1.02 x 10¹⁰ Lyr

Explanation:

Given;

speed of light, v = 300, 000 km/sec

Hubble's constant, H₀ = 2.2 x 10⁻⁵ km/s/Lyr

percentage of the quasar recession = 75% of speed of light

Hubble's Law is given by;

[tex]v =H_od\\\\d = \frac{v}{H_o}\\\\d= \frac{(0.75*300,000)}{2.2*10^{-5}}.Lyr\\\\d = 1.02*10^{10} \ Lyr[/tex]

Therefore, the distance to the quasar is 1.02 x 10¹⁰ Lyr

In this circuit, all the resistors are identical, they are assigned voltage V1, V2 and V3 which correspond to R1 R2 and R3.

Put the volume in order of lowest to highest.

PLEASE HELP, WILL GIVE BRAINLIEST, NEED A GOOD EXPLANATION NOT JUST ANSWER AS IM VERY CONFUSED.

Answers

R2 has the highest Voltage and R3 has the lowest voltage

A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside the solenoid near its center

Answers

Answer:

The value  is  [tex]B =  0.0452 \  T [/tex]

Explanation:

From the question we are told that

   The number of turns is  N  =  1000

    The length is  L =  50 cm =  0.50 m  

    The radius is  r =  2.0 cm  =  0.02 m

     The current is I  =  18.0 A

   

Generally the magnetic field is mathematically represented as

         [tex]B = \mu_o  * \frac{N }{L}  *  I[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space with value  

     [tex]\mu_o  =  4\pi * 10^{-7} N/A^2[/tex]

So

     [tex]B =  4\pi * 10^{-7}  *   \frac{1000}{0.50} *  18.0[/tex]

=>   [tex]B =  0.0452 \  T [/tex]



When driving at slower speeds you need to use what type of steering

wheel movements compared to when driving at faster speeds? *

Answers

Answer:

slower speeds = larger and faster steering wheel movements

faster speeds = small and slow steering wheel movements

Explanation:

When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.

A cart with an unknown mass is at rest on one side of a track. A student must find the mass of the cart by using Newton’s second law. The student attaches a force probe to the cart and pulls it while keeping the force constant. A motion detector rests on the opposite end of the track to record the acceleration of the cart as it is pulled. The student uses the measured force and acceleration values and determines that the cart’s mass is 0.4kg . When placed on a balance, the cart’s mass is found to be 0.5kg . Which of the following could explain the difference in mass?

Answer choices:

A) The track was not level and was tilted slightly downward.

B) The student did not pull the cart with a force parallel to the track.

C) The wheels contain bearings that were rough and caused a significant amount of friction.

D) The motion sensor setting was incorrect. The student set it up so that motion away from the sensor would be the negative direction.

Answers

Answer: The correct answer is A) The track was not level and was tilted slightly downward.

Explanation: This is because of the two values: 0.4 kg and 0.5 kg. I won't go into much detail but due to this difference of mass, we know that the track was not level.

"The track was not level and was tilted slightly downward" could explain the difference in the mass.

Mostly because the university student or learners calculates a mass of just over the spring quantity, the vehicle speed seems to have been higher than there would have had to be.Option B, as well as Option C, are wrong because the acceleration would've been smaller in each of these 2 circumstances, so that computed mass would've been larger.

Thus Option A is appropriate.

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Which is a belief held by sociologists who work from a social-conflict
perspective?
O A. The best approach for a study is from a micro-level orientation.
O B. Personal background has little impact on how individuals react
with one another.
C. Some social patterns are helpful, while others are harmful.
D. Data are irrelevant to the study of sociology.
SUBMIT

Answers

Answer:

C. Some social patterns are helpful, while others are harmful.

Explanation:

Hope this was helpful, Have an amazing,spooky Halloween!!

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the speed of the shell as it hits the ground

Answers

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

which statement is correct about the strength of forces?
-Electrostatic forces are exactly 10 times stronger than gravitational forces.
-Electrostatic forces are exactly 10 times weaker than gravitational forces.
-Electrostatic forces are trillions of times stronger than gravitational forces.
-Electrostatic forces are trillions of times weaker than gravitational forces.

Answers

Answer:

Thanks!!!!! adding this so it doesn’t get deleted.

Explanation:

1. Electrostatic forces are trillions of times stronger than gravitational forces. 2. normal force and friction 3. contact forces 4. The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate. 5. The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.

The correct statement over here is that electrostatic forces are trillions of times stronger than gravitational forces. Hence, option C is correct.

What is an Electrostatic Force?

One of the basic forces in the cosmos is electrostatic force. In the universe, there are four basic forces. These include gravitational force, electromagnetic force, weak nuclear force, and strong nuclear force. Under the umbrella of electromagnetic force is electrostatic force. Two charges placed apart are subject to the electrostatic force. The size of each charged and the separation between them determines how much electrostatic force there will be.

When two charges of the same type are brought together, whether positive or negative, they repel one another. It is known as the electrostatic force of repelling when it operates among two charges that are similar.

Therefore, the electrostatic forces are trillions of times stronger than the gravitational forces.

To know more about Electrostatic Force:

https://brainly.com/question/9774180

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To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .

Answers

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  [tex]  v  =  350 \  m/s [/tex]  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by [tex]k  =  60 cm  =  0.6 \  m[/tex]

    The frequency of both speakers is [tex]f =  700 \  Hz[/tex]

Generally the distance of the listener to the first speaker is mathematically represented as

       [tex]L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}[/tex]

       [tex]L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}[/tex]

        [tex]L_1  =   5.39 \  m [/tex]

Generally the distance of the listener to second speaker at its new position is  

          [tex]L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}[/tex]

       [tex]L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}[/tex]

        [tex]L_2  =   5.64  \  m [/tex]  

Generally the path difference between the speakers is mathematically represented as

        [tex]pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}[/tex]

Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as

         [tex]\lambda =  \frac{v}{f}[/tex]

=>    [tex] L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}[/tex]

=>    [tex] L_2 - L_1  =  \frac{n  *  v}{2f}[/tex]  

=>    [tex] L_2 - L_1  =  \frac{n  *  v}{2f}[/tex]  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    [tex]  5.64 - 5.39   =  \frac{1  *  v}{2*700}[/tex]      

=>    [tex]  v  =  350 \  m/s [/tex]  

The speed of sound in air is 350 m/s

Since the distance between both speakers is 4 m, and the listener is standing 5 m away from halfway between them, the distance L from each loudspeaker to the listener, since the arrangement forms a right-angled triangle, using Pythagoras' theorem,

L = √[(5 m)² + (4/2 m)²]

= √[25 m² + (2 m)²]

= √[25 m² + 4 m²]

= √29 m² = 5.39 m.

Now, when one speaker is moved 60 cm = 0.6 m away from its original position, its distance from the listener is now

L' = √[(5 m)² + (4/2 + 0.6 m)²]

= √[25 m² + (2 m + 0.6 m)²]

= √[25 m² + (2.6 m)²]

= √[25 m² + 6.76 m²]

= √31.76 m²

= 5.64 m.

Now, the path difference when we first have destructive interference is

ΔL = L' - L

= 5.64 - 5.39

= 0.25

Since we have destructive interference for the first time when the speaker is moved, the path difference, ΔL = (n + 1/2)λ where λ = wavelength = v/f where v = speed of sound in air and f = frequency = 700 Hz.

Now, since we have destructive interference for the first time, n = 0.

So,  ΔL = (n + 1/2)λ

ΔL = (0 + 1/2)v/f

ΔL = v/2f

Making v subject of the formula, we have

v = 2fΔL

Substituting the values of the variables into the equation, we have

v = 2fΔL

v = 2 × 700 Hz × 0.25 m

v = 350 m/s

So, the speed of sound in air is 350 m/s

Learn more about interference of sound here:

https://brainly.com/question/1346741

Question #2
2. Stan walks 10km west to the grocery store. He shops
then walks back 10 km east back to his house. What
distance did he cover? What was his displacement?

Answers

Answer:

Distance 20 km and Displacement 0 km

His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.

An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 7.9 times the magnitude of the tangential acceleration. What is the angle

Answers

Answer:

The angle is 3.95 rad.

Explanation:

The angle can be calculated as follows:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]

Where:

[tex]\omega_{f}[/tex]: is the final angular speed

ω₀: is the initial angular speed = 0 (it starts from rest)

α: is the angular acceleration

θ: is the angle=?

The centripetal acceleration is:

[tex]a_{c} = \omega_{f}^{2}*r[/tex]

And the tangential acceleration is:

[tex] a_{T} = \alpha*r [/tex]

Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:

[tex]a_{c} = 7.9a_{T}[/tex]

[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]

Now, the angle is:

[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]

[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]

Therefore, the angle is 3.95 rad.

 

I hope it helps you!          

The angular distance traveled by the electric drill is 3.95 radians.

The given parameters;

initial angular speed, [tex]\omega_i[/tex] = 0centripetal acceleration, [tex]a_c[/tex] = 7.9a

The angular distance traveled by the electric drill is calculated as follows;

[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]

The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;

[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]

Substitute the value of angular acceleration into the first equation;

[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]

Thus, the angular distance traveled by the electric drill is 3.95 radians.

Learn more about angular distance here: https://brainly.com/question/12680957

What is the final velocity (in m/s) of a hoop that rolls without slipping down a 3.50-m-high hill, starting from rest

Answers

Answer:

8.29m/s

Explanation:

Given

Height of hill H = 3.50m

Initial velocity u = 0m/s

Required

Final velocity v

Using the equation of motion

v² = u²+2gH

v² = 0²+2(9.81)(3.5)

v² = 0+7(9.81)

v² = 68.67

v = √68.67

v = 8.29m/s

Hence the final velocity of the hoop is 8.29m/s

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