This is a graph of a car speeding up and then reaching a constant speed of 105 m/s . After it finishes speeding up, how far does the car travel during the last 6 seconds of the graph, from 4 to 10 s?

Answer:

J = 1800 kg-m/s

Explanation:

Given that,

Mass of a boy, m = 150 kg

Initial velocity of a boy, u = 12 m/s

Finally, it stops, v = 0

We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,

[tex]J=m(v-u)\\\\=150\times (0-12)\\\\=-1800\ kg-m/s\\\\|J|=1800\ kg-m/s[/tex]

So, the impulse is equal to 1800 kg-m/s

Answer:

62

Explanation:

it doesn't need explanation

Answer:

A. cause i just took the test

Explanation:

Answer:

its A

Explanation:

no explanation is needed, just trust me.

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]KE = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]KE = \frac{1}{2} \times 2 \times {5}^{2} \\ = 1 \times {5}^{2} [/tex]

We have the final answer as

Hope this helps you

Answer:

The distance the bullet will miss the target is 1.13 m.

Explanation:

Given;

Initial velocity of the bullet = 250 m/s

Distance of the target = 120 m

Time of motion;

t = 120 / 250

t = 0.48 s

During this time the bullet is under the gravitational pull and the distance it will miss the target is given by;

Y = V₀y + ¹/₂gt²

where;

V₀y is the initial vertical velocity = 0

Y = 0+ ¹/₂gt²

Y = ¹/₂(9.8)(0.48)²

Y = 1.13 m

Therefore, the distance the bullet will miss the target is 1.13 m.

Answer: 8 meters per second

Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

f is the force

m is the mass

From the question we have

[tex]a = \frac{100}{85} = \frac{20}{17} \\ = 1.176470...[/tex]

We have the final answer as

Hope this helps you

Answer:

[tex]\mu=0.75[/tex]

Explanation:

Given that,

Force acting on an object, F = 30 N

Mass of the object, m = 4 kg

It is moving with a constant velocity of 2 m/s across a level surface.

We need to find the coefficient of friction  between the object and the surface. Let it is μ. Force in terms of coefficient of friction  is given by :

F = μ N, Where N is normal force, N = mg

[tex]\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{30}{4\times 10}\\\\\mu=0.75[/tex]

So, the coefficient of friction  between the object and the surface is 0.75.

Answer:

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heres your answer

This force of gravitational attraction

is depending on the masses of both

objects and contrarily corresponding

to the square of the distance that divided

their middle parts. This means that as you

move away from an object the gravitational

force lowers. i hope i could help you! ☆

have a great day!

- madeline/madi ✧･ﾟ: *✧･ﾟ:･ﾟ✧*:･ﾟ✧･ﾟ

Explanation:

Answer:

free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it

Explanation:

hope this helped

Answer:A freefall is what you experience when you go skydiving and that is because you are falling out of the sky

Explanation:

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

[tex]\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2[/tex]

(a)

The final angular speed is given as;

[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s[/tex]

(b) the time taken to turn 5.5 revolutions is given as

[tex]\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s[/tex]

Answer:

Explanation:

In thermodynamics, an adiabatic process is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings.Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work. It also conceptually supports the theory used to explain the first law of thermodynamics and is therefore a key thermodynamic concept.

The system takes in heat but does not release it. If an adiabatic process is reversible, there is no change in entropy; otherwise, there is a rise in entropy, or degree of disorder.

What is adiabatic process?

A thermodynamic process known as an adiabatic process takes place when there is no exchange of mass or heat between the system and its surroundings.

An adiabatic process only transfers energy to the environment as work, in contrast to an isothermal process.

The system's overall heat remains constant since there is no heat exchange with the environment.

It is a fundamental idea in thermodynamics and theoretically supports the theory that underlies the first law of thermodynamics.

Therefore, no heat flow in and out of system, which is a characteristic of an adiabatic process.

Learn more about adiabatic process here:

https://brainly.com/question/11938296

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Answer:

The gravitational potential energy of the ball is 13.23 J.

Explanation:

Given;

mass of the ball, m = 0.5 kg

height of the shelf, h = 2.7 m

The gravitational potential energy is given by;

P.E = mgh

where;

m is mass of the ball

g is acceleration due to gravity = 9.8 m/s²

h is height of the ball

Substitute the givens and solve for gravitational potential energy;

PE = (0.5 x 9.8 x 2.7)

P.E = 13.23 J

Therefore, the gravitational potential energy of the ball is 13.23 J.

Explanation:

The dog was stationary at segment c

Answer:

I don't know if I'm wrong but I'm pretty sure stationary means that the thing is still. I would go with C And maybe D????

Answer:

Edwin Hubble found that galaxies are constantly moving away from us. According to his observations with the Hubble Space Telescope, galaxies are moving at different speeds. This shows that the universe is expanding. The farther away a galaxy is, the faster it is moving away. Found this on google hope this helps.

Answer:

A) the universe started at a central point

Explanation:

taking the quiz on eg. :))

An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.

Since the object must be moved away to a distance greater than the radius of the Earth, then change in gravitational potential energy must be based on Newton's Law of Gravitation.

By the Work-Energy Theorem, the work ([tex]W[/tex]), in joules, done on the object is equal to the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:

[tex]W = U_{g}[/tex] (1)

[tex]W = -G\cdot m\cdot M\cdot \left(\frac{1}{r_{f}}-\frac{1}{r_{o}} \right)[/tex] (1b)

Where:

If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 1000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]r_{o} = 6.371\times 10^{6}\,m[/tex] and [tex]r_{f} = 19.113\times 10^{6}\,m[/tex], then the energy required to move the object from the Earth's surface is:

[tex]W = -\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1000\,kg)\cdot (5.972\times 10^{24}\,kg)\cdot \left[\frac{1}{19.113\times 10^{6}\,m} - \frac{1}{6.371\times 10^{6}\,m} \right][/tex][tex]W = 4.171\times 10^{10}\,J[/tex]

An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.

We kindly invite to check this question on gravitational potential energy: https://brainly.com/question/19768887

Answer:

a

 [tex]k = 457333.3 N/m[/tex]

b

[tex]x_a =0.09\ m[/tex]      

Explanation:

From the question we are told that

    The total mass of  three people is  [tex]M = 2.00*10^{2} \ kg[/tex]

     The mass of the car is  [tex]m_c = 1.200 *10^{3} \ kg[/tex]

     The compression of the car spring is  [tex]x = 3 \ cm = 0.03 \ m[/tex]

Generally the spring constant is mathematically represented as

          [tex]k = \frac{F}{x}[/tex]

Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as        

=>       [tex]F = (M +m_c) *g[/tex]

=>       [tex]F = ([2.0*10^{2} ]+[ 1.200*10^{3}]) * 9.8[/tex]

=>       [tex]F = 13720 \ N[/tex]

So

        [tex]k = \frac{13720}{0.03}[/tex]

=>     [tex]k = 457333.3 N/m[/tex]

Generally if the mass which the car is loaded with is  [tex]m = 3.00*10^{2} \ kg[/tex]

Then the force experienced by the spring is  

         =>       [tex]F_a = (m +m_c) *g[/tex]

         =>       [tex]F_a = (3.00*10^{3} + 1.200 *10^{3}) * 9.8[/tex]

         =>       [tex]F_a = 41160 \ N[/tex]

Generally from the above formula the compression is  

       [tex]x_a = \frac{F_a}{k}[/tex]

=>    [tex]x_a = \frac{41160}{457333.3}[/tex]

=>    [tex]x_a =0.09\ m[/tex]      

Answer:

True

Explanation:

have a good day:)

Answer: This statement is True

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

Answer:

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Explanation:

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Answer:

Magnitude of the induced emf is 11.62 V

Explanation:

Given:

No. of turns of the circular coil, N=185

Radius, R=4.50 cm=0.045 m

No. of turns of solenoid per meter(m), n=350

a=12.5 A

b=2.10 [tex]s^{-1}[/tex]

Now,

To determine the emf induced in the coil at t = 1.50 s:

The given equation is:

[tex]I=a(1-e^{-bt})[/tex]

Now,

[tex]B=\mu_{o}nI=\mu_{o}na(1-e^{-bt})[/tex]

[tex]\frac{dB}{dt}=\mu_{o}nabe^{-bt}[/tex]

Now, substituting the respective values:

[tex]\frac{dB}{dt}=-4\pi\times 10^{-7}\times 350\times 12.5\times 2.10e^{-2.10\times 1.50}[/tex]

[tex]\frac{dB}{dt}=0.988\ T/s[/tex]

Now,

[tex]emf=NA\frac{dB}{dt}[/tex]

where,

A=Area=[tex]\pi R^{2}[/tex]

Thus,

[tex]emf=185\times \pi\times (4.50\times 10^{-2})^{2}\times 11.545e^{-3.15}[/tex]

[tex]emf=11.62 V[/tex]

Answer:

[tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]

[tex]6.178\times 10^{-34}\ \text{m/s}[/tex]

[tex]0.31\times 10^{-4}\ \text{m/s}[/tex]

Explanation:

[tex]\Delta x[/tex] = Uncertainty in position = 1.9 m

[tex]\Delta p[/tex] = Uncertainty in momentum

h = Planck's constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]

m = Mass of object

From Heisenberg's uncertainty principle we know

[tex]\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}[/tex]

The minimum uncertainty in the momentum of the object is [tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]

Golf ball minimum uncertainty in the momentum of the object

[tex]m=0.045\ \text{kg}[/tex]

Uncertainty in velocity is given by

[tex]\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}[/tex]

The minimum uncertainty in the object's velocity is [tex]6.178\times 10^{-34}\ \text{m/s}[/tex]

Electron

[tex]m=9.11\times 10^{-31}\ \text{kg}[/tex]

[tex]\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}[/tex]

The minimum uncertainty in the object's velocity is [tex]0.31\times 10^{-4}\ \text{m/s}[/tex].

Answer:

The speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.

Explanation:

The radial or centripetal acceleration is given by:

[tex] a_{c} = \frac{v^{2}}{r} [/tex]

Where:

v: is the speed = v₀

r: is the radius = r₀

[tex] a_{c} = \frac{v_{0}^{2}}{r_{0}} [/tex]    (1)

If the radius is now equal to half the initial radius the speed must be:

[tex]a_{c} = \frac{v^{2}}{r_{0}/2}[/tex]    (2)

By equating equation (1) and (2):

[tex] \frac{v_{0}^{2}}{r_{0}} = \frac{v^{2}}{r_{0}/2} [/tex]  

[tex]v^{2} = \frac{v_{0}^{2}}{2}[/tex]

[tex] v = \frac{v_{0}}{\sqrt{2}} [/tex]      

Therefore, the speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.      

I hope it helps you!                      

Answer:

[tex]t=0.0107\ \text{s}[/tex]

Explanation:

[tex]x=10\cos(6t)[/tex]

Now [tex]x=8\ \text{cm}[/tex]

Substituting the value of [tex]x[/tex] in the equation we get

[tex]8=10\cos6t[/tex]

[tex]\Rightarrow 0.8=\cos6t[/tex]

[tex]\Rightarrow \cos^{-1}0.8=6t[/tex]

[tex]\Rightarrow t=\dfrac{\cos^{-1}0.8}{6}[/tex]    the values here are used found in radians

[tex]\Rightarrow t=0.0107\ \text{s}[/tex]

So, at [tex]t=0.0107\ \text{s}[/tex] the value of [tex]x=8\ \text{cm}[/tex].

Answer:

answer is 11.76 meter

Explanation:

use 2nd equation of motion

S=ut+1/2at^2

Answer:

19.01 N

Explanation:

F = Force being applied to the crate = 45 N

[tex]\theta[/tex] = Angle at which the force is being applied = [tex]65^{\circ}[/tex]

Horizontal component of force is given by

[tex]F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}[/tex]

The horizontal component of the force acting on the crate is 19.01 N.