When the air resistance can be ignored the velocity of an object dropped initially from rest is given by the following equation where g is free-fall acceleration

A. g*t^2/2
B. g*t
C. g*t/2
D. g

Answer:

4 bobux

Explanation:

one bobux

two bobux

three bobux

four bobux

Complete question is;

Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at

(x, y) = (0, 0) and that Allyson and Adrian move in the positive y and negative x directions, respectively. Let one unit equal one foot.)

Compute the length of the bungee cord at t = 7 seconds. (Round your answer to three decimal places.)

Answer:

Length of bungee cord = 85.734 ft

Explanation:

We are told that Adrian moves 9ft/sec. Thus, at 5.5 seconds, distance he moved is; 9 ft/sec × 5.5sec = 49.5 ft in the negative x (-x) direction. Therefore, the coordinate is (-49.5, 0).

Now, Allyson has moved 10ft/sec. Thus, at 7 seconds, distance he moved would be; 10 ft/sec x 7sec = 70 feet in the positive (+y) direction. Therefore, the coordinate is (0, 70).

Now, since they started from the origin, it means (0, 0) is a coordinate. Thus, we now have 3 coordinates which are; (0, 0), (0, 70) & (-49.5,0). These 3 coordinates would therefore combine to form a right triangle.

The hypotenuse is the distance between Allyson and Adrian.

Thus, from pythagoras theorem, we can find the distance between them which is same as the length of the cord.

Thus;

(-49.5)² + 70² = D².

D² = 2450.25 + 4900

D = √7350.25

D = 85.734 ft

Answer:

True.

Explanation:

        [tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)

        [tex]\Delta x} = v_{avg}* {\Delta t}[/tex]

Answer:

Geocentric model of the solar system helped to explain retrograde motions of planets. Explanation: Geocentric model of planets was proposed by Ptolemy. It stated that all sun, planets and stars revolve round the earth in circular orbits.

Answer:

retrograde motion

Explanation:

i reverse searched the image

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

Answer:

1.152 miles

Explanation:

Given: central angle = 1 minute = [tex](\frac{1}{60}) ^{o}[/tex]

           radius of the earth = 3960 miles

The length of an arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r

where: [tex]\alpha[/tex] is the central angle, and r is the radius.

Thus,

Distance along the arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r

Distance along the arc = [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 2 x [tex]\frac{22}{7}[/tex] x 3960

                                      = [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 24891.4286

                                      = 1.1524

The required distance along an arc is 1.152 miles.

Answer:

D

Explanation:

Answer:

c.4s

Explanation:

Answer:

Explanation:

The work done by an object can be found by using the formula

From the question

force = 290 N

distance = 5 m

We have

workdone = 290 × 5

We have the final answer as

Hope this helps you

Answer:

In two significant figure 360K

Explanation:

The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.

Hence,

ΔT = 212 - 32

ΔT = 180°F

To convert to °F to kelvin, we use the formula below

= (°F - 32) × 5/9 + 273.15

= (180°F - 32) × 5/9 + 273.15

= 355.37K ⇔ 360K

Answer:

A mixture is a combination of two or more substances in any proportion. This is different from a compound, which consists of substances in fixed proportions. ... The lemonade pictured above is a mixture because it doesn't have fixed proportions of ingredients.

Explanation:

Answer:

The region where an electron is most likely to be is called an orbital. Each orbital can have at most two electrons. Some orbitals, called S orbitals, are shaped like spheres, with the nucleus in the center.

Explanation:Hope this helps :)

By definition, a region around the nucleus of an atom where electrons are likely to be found​ is called an orbital.

First of all, an atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element.

All atoms are made up of subatomic particles: protons and neutrons, which are part of their nucleus, and electrons, which revolve around them. Protons are positively charged, neutrons are neutrally charged, and electrons are negatively charged.

In other words, every atom consists of a nucleus in which neutrons and protons meet and energy levels where electrons are located.

This is, the atomic nucleus is the central part of the atom that is made up of protons and neutrons, while the orbitals or peripheral region is an area where electrons are found.

In summary, a region around the nucleus of an atom where electrons are likely to be found​ is called an orbital.

Learn more:

Answer:

v₁f = -6.2 cm/s

Explanation:

        [tex]m_{1} *v_{1o} = m_{1}* v_{1f} + m_{2}* v_{2f}[/tex]

       [tex]\frac{1}{2}*m_{1}*v_{1o}^{2} = \frac{1}{2}*m_{1}*v_{1f} ^{2} + \frac{1}2}*m_{2}*v_{2f}^{2}[/tex]

       [tex]v_{1f} = \frac{-m_{1} }{3*m_{1}} *v_{1o} =\frac{-v_{1o}}{3} = -\frac{18.6 cm/s}{3} = -6.2 cm/s[/tex]

Answer:

a- More heat is required for the constant-pressure process than for the constant-volume

Explanation:

we have to solve using the thermodynamic first law. this is the heat applied to the system

dQ = dU + dW

definition of terms:

dU = change in internal energy

dW = work done

we have it that

change in internal energy dU is directly proportional to work done dW

but when we are in constant volume process, work done of the gas is zero

therefore

dQ of constant pressure is > than that of constant volume

so constant pressure process requires more heat

The process that requires more heat is the constant-pressure process than the constant-volume process.

According to the first law of thermodynamics, the heat that's applied to the system will be the addition of the change in internal energy and the work done.

In a constant-volume process, the work done on the gas is equal to zero. More heat will be required for the constant-pressure process than for the constant-volume process.

Also, it should be noted that the change in the thermal energy of the gas will be the same for the constant-pressure process and the constant-volume process.

Read related link on:

https://brainly.com/question/16951562

Answer:

hi

Explanation:

hiijjjjjjjjjjjjjjj

Answer:

6000 joules

Explanation:

I jus learned dis

Answer:6000j

Explanation:

Hope that helps

Answer:

luster

Explanation:

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  B

Explanation:

   The number  of alphabet is  n= 4  (a , b , c , d )

Generally the total  number of  string of length 10 over the 4 alphabets is  

     [tex]N  =  4^{10}[/tex]

Gnerally the number of string of length 10 that does not include b is  

     [tex]T =  3^{10}[/tex]    

Generally the number of strings of length 10 over the 4  alphabets that have at least one alphabet b  somewhere in the string is  

        [tex]G  =  N - T[/tex]

=>    [tex]G  =  4^{10} -  3^{10}[/tex]

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;

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.

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Answer:

[tex]\mu_{s} = 3.071[/tex]

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

Explanation:

From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:

[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)

[tex]\Sigma F_{y} = N-m\cdot g = 0[/tex] (Eq. 2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

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

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

[tex]v[/tex] - Speed of the motorcyclist, measured in meters per second.

[tex]R[/tex] - Radius of the circular turn, measured in meters.

The static coefficient of friction is cleared in (Eq. 1):

[tex]\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}[/tex]

From (Eq. 2) we get that normai force is:

[tex]N = m\cdot g[/tex]

And we expand the resulting expression in (Eq. 1):

[tex]\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}[/tex]

[tex]\mu_{s} = \frac{v^{2}}{g\cdot R}[/tex] (Eq. 3)

If we know that [tex]v = 30.556\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 31\,m[/tex], the expected static coefficient of friction is:

[tex]\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}[/tex]

[tex]\mu_{s} = 3.071[/tex]

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

Answer:

A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.

Explanation:

Answer:

Hyperbole

Explanation:

this is an extreme exaggeration or overstatement/ magnification

Answer and Explanation:

TL: DR The Sun is much more massive than Neptune — more than enough to make up for the somewhat smaller distance between the two planets at the closest approach.

[The surprise in this answer (to me, a non-astronomer), is that the gap between the orbits of Neptune and Uranus is large — half the distance from Uranus to the Sun.]

The ratio of gravitational attraction of the Sun on Uranus versus Neptune on Uranus is directly proportional to the ratio of the Sun’s mass to Neptune’s and inversely proportional to the ratio of the square of the distances (let’s use the closest approach of the two planets to one another to calculate a maximum attraction).

Numbers:

Sun’s mass: 2 x 10^30 kg

Neptune’s mass: 1 x 10^26 kg

Distance of Sun to Uranus: 3 x 10^9 km

Closest approach of Uranus and Neptune: 1.5 x 10^9 km

Without doing any arithmetic, we see that even at their closest approach, Uranus and Neptune are separated by about one-half of the Uranus to Sun distance. Squaring that ratio, we see that if the Sun and Neptune had the same mass, the attraction between the Sun and Uranus would only be about 1/4 of that between the Sun and Neptune; however, the Sun has 20000 times the mass of Neptune, so the attraction between Uranus and the Sun is about 5000 times stronger than the maximum attraction between Uranus and Neptune.

The explanation of the possibility of why sun exerts a greater force on Uranus than Neptune exerts on Uranus is; because the force was calculated to be greater.

The formula for calculating the Force of Gravity between two masses is:

F = G*m₁*m₂/r²

Where;

F = force of gravity

G = gravitational constant = 6.674 × 10⁻¹¹ N•m²/kg²

m₁ = mass of the larger object

m₂ = mass of the smaller object

r = the distance between the centers of the two masses

Now, from online values, we have the following;

mass of Neptune; m₁ =  102.413 × 10²⁴ kg

mass of Uranus; m₂ = 86.813 × 10²⁴ kg

average distance between the centers of Neptune and Uranus; r = 1.62745 × 10¹² m

Thus, force exerted by Neptune on Uranus is;

F = (6.674 × 10⁻¹¹ × 102.413 × 10²⁴ × 86.813 × 10²⁴)/(1.62745 × 10¹²)²

F = 2.240 × 10¹⁷ N

We are told that the force the Sun exerts on Uranus is 1.41 the force the Sun exerts on Uranus is 1.41 × 10²¹ N.

That is greater than the force Neptune exerts on Uranus.

Read more about Force of Gravity at; https://brainly.com/question/7281908

Answer:

d. kT ln 3

Explanation:

Given;

k as Boltzmann constant

Let initial initial microstate, = Фi

Let the final microstate, = Фf = 3Фi

at constant temperature, T

The energy input to gas as heat is given by;

Q = TΔS = Tk(lnФf - lnФi)

[tex]Q= kT*\frac{ln \phi _f}{ln \phi_i} \\\\Q= kT*\frac{3ln \phi _i}{ln \phi_i}\\\\Q= kT ln3[/tex]

Therefore, the energy input to gas as heat is kT ln 3

Answer:

M V = m v        conservation of momentum (Caps-cannon  Small-projectile)

V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s    recoil velocity of cannon

KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2  = 40 kg m^2/s^2 = 40 J

Answer:

(a)  24.56 N

(b) 142.28 N

Explanation:

(a)

The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.

The centripetal function is generated from either scenario by Equation:

⇒  [tex]Fc = \frac{mv^2}{r}[/tex]

On putting the values, we get

⇒       [tex]=\frac{12\times 1.33^2}{0.864}[/tex]

⇒       [tex]=24.56 \ N[/tex]

(b)

Use T to denote whatever arm stress we can get at the bottom including its circle:

⇒  [tex]Fc = T - mg =\frac{ mv^2}{r}[/tex]

⇒  [tex]T = mg + Fc[/tex]

⇒      [tex]=12\times 9.81+24.56[/tex]

⇒      [tex]=142.28 \ N[/tex]