1 An object is thrown vertically upward with an initial speed of 30m/s. after 4s the object is:
A moving down at 20m/s

B. moving up at 9.2 m/s
C. moving up at 9.2m/s
Dat its maximum height
E moving down at 9.2 m/s​

Answer:

answer is 11.76 meter

Explanation:

use 2nd equation of motion

S=ut+1/2at^2

Answer:

  t = 0.6933 s

Explanation:

This is a rotational kinematics exercise, let's find the angular acceleration of the wheel

          w² = w₀² + 2 α θ

          α = (w² - w₀²) / 2 θ

Let's reduce the angles to the SI system

         θ  = 16 rev (2π rad / 1 rev) = 32π rad

let's calculate

          α = (105² - 185²) / (2 32π)

          α = -115.39 rad / s²

now let's use the relation

           w = w₀ + α t

           t = (w- w₀) /α

           t = (105 - 185) / (- 115.39)

           t = 0.6933 s

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:

True

Explanation:

have a good day:)

Answer: This statement is True

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

[tex]N_p= 500 \ turn[/tex]

[tex]N_s= 200 \ turn[/tex]

[tex]I_s= 3.0 \ A[/tex]

By using the transformer formula, we get

⇒ [tex]\frac{N_p}{N_s} =\frac{I_s}{I_p}[/tex]

⇒ [tex]I_p = I_s\times \frac{N_s}{N_p}[/tex]

On substituting the given values, we get

⇒      [tex]=3.0 \ A\times \frac{2000}{500}[/tex]

⇒      [tex]=12.0 \ A[/tex]

Answer:

The angular displacement of the blade is 576,871.2 radians

Explanation:

Given;

angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)

time of motion, t = 3 hours

The angular speed of the Helicopters rotor blades in radian per second is given as;

[tex]\omega = \frac{510 \ rev}{mins} *\frac{2 \pi \ rad}{1 \ rev} *\frac{1 \ min}{60 \ s}\\\\\omega = 53.414 \ rad/s[/tex]

The angular displacement in radian is given as;

θ = ωt

where;

t is time in seconds

θ = (53.414)(3 x 60 x 60)\\

θ = 576,871.2 radians

Therefore, the angular displacement of the blade is 576,871.2 radians

Answer:

A. cause i just took the test

Explanation:

Answer:

its A

Explanation:

no explanation is needed, just trust me.

Answer:

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Explanation:

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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:

[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].

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.

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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:

♡ madeline here ♡

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:

Work done is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].

Explanation:

The work done by the spring is the same as the potential energy stored in the spring.

So that,

work done = potential energy = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]

where k is the spring constant of the material of the spring, and x is the compression.

When the spring is compressed by x;

work done = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]

When the spring is compressed by 2x;

work done = [tex]\frac{1}{2}[/tex] k[tex](2x)^{2}[/tex]

                  = [tex]\frac{1}{2}[/tex] k(4[tex]x^{2}[/tex])

                  = 2k[tex]x^{2}[/tex]

Therefore,

The work done the second time more than the first = 2k[tex]x^{2}[/tex] - [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]

                                                                                = [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex]

The work done the second time more than the first is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].

Spring work is equivalent to the effort done to extend the spring, Work did you do the second time than the first time will be  [tex]\rm \frac{3}{2} Kx^2[/tex].

Spring work is equivalent to the effort done to extend the spring, which is dependent on both the spring constant k and the distance stretched.

The potential energy stored as a result of the deformation of an elastic item, such as spring stretching, is referred to as elastic potential energy.

Work done by spring = potential energy

[tex](PE)_{spring }= \frac{1}{2} Kx^2[/tex]

Case 1

spring is compressed by x

[tex](PE)_1{spring }= \frac{1}{2} Kx^2[/tex]

Case 2

spring is compressed by 2x

[tex]\rm (PE)_2{spring }= \frac{1}{2} K(2x)^2\\\\\rm (PE)_2{spring }= \frac{1}{2}\times 4K(x)^2\\\\\rm (PE)_2{spring }= 2K(x)^2[/tex]

The difference in the potential energy is found by;

[tex](PE)_2-(PE)_1=2Kx^2-\frac{1}{2} Kx^2=\frac{3}{2} Kx^2[/tex]

Hence spring work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].

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https://brainly.com/question/3317535

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. :))

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:

[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].

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:

62

Explanation:

it doesn't need explanation

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:

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.

Answer:

Option 3: -48 cm

Explanation:

We are given:

refractive index; n = 1.5

radius of curvature; r2 = 24 cm

Formula for the focal length is given as;

1/f = (n - 1) × [(1/r1) - (1/r2)]

As r1 tends to infinity, 1/r1 = 0

Thus,we now have;

1/f = (n - 1) × (-1/r2)

Plugging in the relevant values;

1/f = (1.5 - 1) × (-1/24)

1/f = -0.02083333333

f = -1/0.02083333333

f = -48 cm

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:

[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:

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!