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]
The scientific method is the only way of learning about Nature used by scientist today *
A. true
B. false
Answer:
false
Explanation:
The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The original voltage is [tex]V_o[/tex]
The new voltage is [tex]V =\frac{V_o}{2}[/tex]
The capacitance is [tex]C = 150\ nF = 150 *10^{-9} \ F[/tex]
The first resistance is [tex]R_i = 26 \Omega[/tex]
The second resistance is [tex]R_E = 200 \Omega[/tex]
Generally the equivalent resistance is
[tex]R_e = R_1 + R_E[/tex]
=> [tex]R_e = 26 +200 [/tex]
=> [tex]R_e = 226 \ \Omega [/tex]
Generally the time constant is mathematically represented as
[tex]\tau = RC[/tex]
=> [tex]\tau = 226 * 150 *10^{-9}[/tex]
=> [tex]\tau = 3.39 *10^{-5} \ s [/tex]
Generally the voltage is mathematically represented as
[tex]V = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]\frac{V_o}{2} = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]0.5 = e^{-\frac{t}{\tau} }[/tex]
=> [tex]ln(0.5) = {-\frac{t}{ 3.39 *10^{-5} } }[/tex]
=> [tex]ln(0.5) * 3.39 *10^{-5} = -t [/tex]
=> [tex]t = 2.35*10^{-5} \ s [/tex]
On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)
Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]
Therefore, the height of the cliff is 121.276 m
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.
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 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.
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
How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 290 N?
Answer:
The answer is 1450 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
force = 290 N
distance = 5 m
We have
workdone = 290 × 5
We have the final answer as
1450 JHope this helps you
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.
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.
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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 .
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
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Is it true or false that the displacement always equals the product of the average velocity and the time interval?
Answer:
True.
Explanation:
Applying the definition of average velocity, we know that we can always write the following expression:[tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)
By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:[tex]\Delta x} = v_{avg}* {\Delta t}[/tex]
as a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is
Answer:
The value of q is [tex]\dfrac{Q}{8}[/tex]
Explanation:
Given that,
Each charge = -Q
Distance between charges = L
Reduced force = [tex]\dfrac{F}{2}[/tex]
Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.
We know that,
The force on each end is
[tex]F=\dfrac{kQ^2}{L^2}[/tex]...(I)
If the charge q is placed at mid point then
The force on each end charge is
[tex]\dfrac{F}{2}=F+F'[/tex]....(II)
We need to calculate the value of q
Using equation (II)
[tex]\dfrac{F}{2}=F+F'[/tex]
Put the value of F into the formula
[tex]\dfrac{\dfrac{kQ^2}{L^2}}{2}=k\dfrac{Q^2}{L^2}+k\dfrac{q\times(-Q)}{(\dfrac{L}{2})^2}[/tex]
[tex]\dfrac{kq(-Q)}{(\dfrac{L}{2})^2}=-\dfrac{kQ^2}{2L^2}[/tex]
[tex]\dfrac{q}{\dfrac{1}{4}}=\dfrac{Q}{2}[/tex]
[tex]q=\dfrac{Q}{8}[/tex]
Hence, The value of q is [tex]\dfrac{Q}{8}[/tex]
6) The magnitude of the force the Sun exerts on Uranus is 1.41 x 1021 newtons. Explain how it is possible for the Sun to exert agreater force on Uranus than Neptune exerts on Uranus.
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.
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Plz help me fast WITH EXTRA POINTS AFTER SUBMITTING
Answer:
4 bobux
Explanation:
one bobux
two bobux
three bobux
four bobux
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.
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
An ideal gas increases in temperature from 22°C to 42°C by two different processes. In one process, the temperature increases at constant volume, and in the other process the temperature increases at constant pressure. Which process requires more heat or are the required amount of heat same in both?
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.
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How do compounds differ from mixtures such as lemonade
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:
The word acid comes from the Latin word
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?
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
Find the work done by a 75.0 kg person in climbing a 2.50 m flight of stairs at a constant speed.
Answer:
1,839.375 Joules
Explanation:
Work is said to be done is the force applied to an object cause the object to move through a distance.
Workdone = Force * Distance
Workdone = mass * acceleration due to gravity * distance
Given
Mass = 75.0kg
acceleration due to gravity = 9.81m/s²
distance = 2.50m
Substitute the given parameters into the formula:
Workdone = 75.0*9.81*2.50
Workdone = 1,839.375Joules
Hence the workdone is 1,839.375 Joules
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
Answer:
D. 10 m/s
Explanation:
A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid cylindrical hub of mass 0.0055 kg and diameter 0.011 m . Part A Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest. Express your answer using three significant figures and include the appropriate units.
Answer: IM 95%sure that the answer is B jus took the test got the answer right
Explanation:
Answer:
sorry I forgot I wish I could help
What are the standard international (si) units of distance
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.
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?
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
for an emitted wavelength of 500 nanometers and a redshift of 0.4 what will be the observed wavelength g
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]
Time it takes stone to fall from the height of 80 m is approximately equal to *
A. 1 s
B. 2 s
C. 4 s
D. 8 s
Answer:
D
Explanation:
Answer:
c.4s
Explanation:
my heart strike him to dead.what figure of speech is that?
Answer:
Hyperbole
Explanation:
this is an extreme exaggeration or overstatement/ magnification
See Conceptual Example 6 to review the concepts involved in this problem. A 12.0-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 86.4 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 1.33 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.
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]
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?
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.
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
Answer:
C. Some social patterns are helpful, while others are harmful.
Explanation:
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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
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
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