Two carts of equal mass are on a horizontal surface with negligible friction. Cart A is approaching cart B, which is at rest, as shown in Figure 1 above. Attached to cart B is a spring that is initially compressed. At the moment cart A collides with cart B, the spring is released and pushes on cart A, as shown in Figure 2 above. Which of the following correctly states what happens to the kinetic energy and the momentum of the two-cart system as a result of the collision compared to those quantities before the collision?
(A) Kinetic energy: Increases; Magnitude of Momentum: Decreases
(B) Kinetic energy: Increases; Magnitude of Momentum: Stays the same
(C) Kinetic energy: Increases; Magnitude of Momentum: Increases
(D) Kinetic energy: Stays the same; Magnitude of Momentum: Decreases
(E) Kinetic energy: Stays the same; Magnitude of Momentum: Stays the same

The displacement, acceleration, energy and velocity of the simple harmonic motion of the mass attached to the spring are as follows;

A) x(t) = 0.05·sin(10·t + π/2)

B) The acceleration is; a(t) = -2.5 m/s²

C) The total energy is 0.0625 J

D) The velocity is ±√3/4 m/s

The restoring force of a body in simple harmonic motion is directly proportional to the displacement of the body from its mean or central position.

Mass of the block, m = 2 kg

The spring constant, k = 200 N/m

The extension of the spring = 5 cm

Time at which the spring is released, t = 0

A. The motion of the spring with the mass is a Simple Harmonic Motion

The angular velocity can be obtained using the formula;

ω = √(k/m)

Therefore;
ω = √(200/2) = 10

The angular velocity of the block on the spring is, ω = 10 rad/s

The period, T = The time to complete 2·π rad

Therefore; T = 2·π rad/(10 rad/s) = π/5 s

The amplitude, A, is the cistance of the mass from the at rest position, which is 5 cm = 0.05 m

The equation of the extension of the spring is therefore;

x(t) = 0.05·sin(10·t + c)

At t = 0, x(t) = 0.05, therefore;

sin(10 × 0 + c) = sin(c) = 1

c = π/2

The equation for the displacement as a function of time is therefore;

B. The acceleration when x(t) = A/2 is obtained as follows;

x(t) = 0.05·sin(10·t + π/2)

A/2 = 0.05·sin(10·t + π/2)

A = 0.05

0.05/2 = 0.05·sin(10·t + π/2)

sin(10·t + π/2) = 1/2

10·t + π/2 = π/6

t = -π/30

cos(10×(-π/30) + π/2) = ±√3/2

v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)

a(t) = v'(t) = -5·sin(10·t + π/2)

a(t) = v'(t) = -5·sin(10·t + π/2) = -5 × 1/2 = -2.5

C. The energy in a pring = (1/2)·k·x²

When x = A/2, we get;

E = (1/2) × 200 × (0.05/2)² = 0.0625

D) The velocity when x = A/2 is; v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)

v(t) = 0.5 × cos(10·t + π/2)

When x = A/2, sin(10·t + π/2) = 1/2, therefore;

cos(10·t + π/2) = ±√3/2

v(t) = 0.5 × ±√3/2 = ±√3/4

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Answer:

70.56 J

Explanation:

Gravitational Potential Energy= mass×gravitational pull× height

= 6×9.8×1.2= 70.56 J

Answer:

v3 = 0 gm/s / 700g

Explanation:

To solve this problem, you need to use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the total momentum of the system (the 2kg object before it explodes) is equal to the sum of the momenta of the three pieces after the explosion.

You can calculate the momentum of each piece by multiplying its mass by its velocity:

P1 = 800g * 30 m/s = 24,000 gm/s

P2 = 500g * 520 m/s = 260,000 gm/s

The total momentum of the system is the sum of these two momenta:

Ptotal = P1 + P2 = 24,000 gm/s + 260,000 gm/s = 284,000 g*m/s

The third piece has a mass of 2kg - 800g - 500g = 700g. We can use the conservation of momentum equation to find its velocity:

Ptotal = (700g * v3) + (800g * 30 m/s) + (500g * 520 m/s)

v3 = (Ptotal - (800g * 30 m/s) - (500g * 520 m/s)) / 700g

v3 = (284,000 gm/s - (800g * 30 m/s) - (500g * 520 m/s)) / 700g

v3 = (284,000 gm/s - 24,000 gm/s - 260,000 gm/s) / 700g

v3 = (284,000 - 24,000 - 260,000) gm/s / 700g

v3 = 0 gm/s / 700g

The velocity of the third mass is 0 m/s.

Hope this helps.

Answer:

d = 5 m

Explanation:

In this exercise we have a graph of displacement against time, the graph being a line, so the body has a uniform movement, the speed of the person is

         v = [tex]\frac{\Delta x}{\Delta t}[/tex]

         v =[tex]\frac{5-0}{5-0}[/tex] (5-0) / (5-0)

         v = 1 m / s

therefore the displacement is

         d = v t

         d = 1  5

          d = 5 m

To find the normal force, we can use the equation: normal force = weight + friction force * cos(incline angle).

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Answer:

  F = 3.6 10⁶ N

Explanation:

The expression for the electric force is

         F =  [tex]k \ \frac{q_1 q_2}{r^2}[/tex]

in this case it indicates that the charge q1 is doubled

       q₁ = 2   3 10⁻³ C

       q₁ = 6 10⁻³ C

let's reduce the magnitudes to the SI system

        q₂ = 6 10⁻³ C

        r = 0.3 m

let's calculate

       F = 9 10⁹ 6 10⁻³ 6 10⁻³ / 0.3²

       F = 3.6 10⁶ N

Total distance covered by Brandon is 18 km and total displacement covered by him is 13.41 km.

The definition of displacement is the changing of an object's position. It has a magnitude and direction and is a vector quantity. It is shown as an arrowhead that travels from the initial location to the end. An object's position changes, for instance, if it moves from position A to position B.

Distance covered by Brandon is-

12 + 6 = 18 km

Displacement covered by Brandon is-

d²= 12² + 6²

= 144 + 36

d²= 180

d = √180

d = 13.41 km.

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Answer: e. Christian Dopplerâ

Explanation:

Based on the information given, the scientist of the past that should definitely be included in the exhibit is Christian Dopplera.

He described how the frequency of sound waves and light is being affected by the relative speed of both the source and also the observer. This was referred to as the Doppler effect.

In this scenario, the Doppler effect can be used to show how the universe is expanding. Therefore, the correct option is E.

Answer:

Choice A. gamma rays

Explanation:

The displacement of the iguana between 3 s and 6 s  is 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s is  8.08 meters.

Distance is the sum of an object's movements, regardless of direction.

The term "displacement" refers to a shift in an object's position.

According to the graph:

The displacement of the iguana between 3 s and 6 s

= √{ (3-6)²+(6-0)²} meters

= 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s

= [ √{ (3-5)²+(6-6)²} +√{ (5-6)²+(6-0)²}] meters

= [2+ 6.08] meters

= 8.08 meters.

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Answer:

v = 2 m/s

Explanation:

Here, we will use the law of conservation of momentum to solve this problem:

[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of grandma = 80 kg

m₂ = mass of James = 40 kg

u₁ = initial speed of grandma = 3 m/s

u₂ = initial speed of James = 0 m/s

v₁ = v₂ = v = final speed of grandm and James = ?

Therefore,

[tex](80\ kg)(3\ m/s)+(40\ kg)(0\ m/s)=(80\ kg)(v)+(40\ kg)(v)\\\\(120\ kg)v = 240\ Ns\\\\v = \frac{240\ N.s}{120\ kg}\\[/tex]

v = 2 m/s

Answer:

As a bat increases its forward speed, it should decrease the frequency of the emitted pulses to compensate.

decrease

Explanation:

Decreasing the frequency of the emitted pulse will help the bat reduce its frequency caused by its forward motion.  The forward motion shifts the bat's auditory frequency to a higher frequency; consequently, the bat should adjust downwards the frequency of the emitted pulse so the reflected pulse will be in the correct frequency range.

Answer:

Explanation:

When a yellow filter is used, the stage light will appear yellow because the filter absorbs all colors of light except for yellow, which it allows through. When an orange filter is used, the stage light will appear orange for the same reason.

It is important to note that the color of the lamp itself will not change, but the light that is seen by the observer will appear to be the color of the filter due to the absorption and transmission properties of the filte

Answer:

Explanation:

The period of oscillation will remain unchanged because the period of oscillation of a pendulum does not depend upon the mass of the bob  . Here monkey along with bunch of banana represents bob .

When the monkey and banana were at height h /2 , they have potential energy as well as kinetic energy . banana is separated from the system . It carried its total energy along with it . But the energy of monkey remained intact with it . So it will keep on moving as usual . So it will attain the same maximum height as before .

Hence the amplitude of oscillation too will remain unchanged .

Answer:

Radioactive materials give off a form of energy called ionizing radiation.

Answer:

0.5 seconds

Explanation:

Work = Fdcos(theta) = 10*2*cos(0) = 20 J

Power = W/t

40 = 20/t

t = 0.5s

Answer:

did any of this help

Explanation:

y = (-2/3)x - 1

y-(-5)= -2/3(x-6)

y-y1=m(x-x1)

2x-3y=11

Capsule communicator or Capcom  is the only group in mission control that gets to communicate with the astronauts in space. Hence, option (B) is correct.

The capsule communicator, or Capcom, was the only voice that spoke to the astronauts during their trip to avoid any mistake. To ensure that the men in the capsule always had a familiar individual who understood their situation and could provide the information they required, Capcom was always manned by astronauts.

Both the technical control team on the ground and the astronauts in space are represented by Capcom.

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Answer:

a)   v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex],   b)  Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex], c)  T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]

Explanation:

a) For this exercise we must use Newton's second law with the gravitational force

          F = ma

          [tex]G \frac{m M_e}{(R+R_e)^2 }[/tex] = m a

the acceleration of the satellite is centripetal

          a = [tex]\frac{v^2}{(R+R_e)}[/tex]

we substitute

            [tex]G \frac{m M_e}{(R+R_e)^2 } = m \frac{v^2}{ (R+R_e)}[/tex]

          [tex]G \frac{M_e}{(R+R_e)}[/tex]  = v²

          v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex]

the distance is from the center of the earth

b) mechanical energy is the sum of kinetic energy plus potential energy

         Em = K + U

         Em = ½ m v² - G m M / (R + R_e)

we substitute the expression for the velocity

         Em = ½ m  [tex]G \frac{M_e}{(R+R_e)}[/tex]  - [tex]G \frac{M_e}{(R+R_e)}[/tex]  

         Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex]  

c) as the orbit is circulating, the velocity modulus is constant

         v = d / t

in a complete orbit the distance traveled of the circle is

        d = 2π (R + R_e)

where time is called period

         v = 2π (R + R_e)

         T = 2π (R + R_e) / v

we substitute the speed value

        T = 2π (R + R_e) [tex]\sqrt{\frac{(R+R_e) }{G M_e } }[/tex]

        T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]

(a) An expression for the speed of the satellite in its orbit.

[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]

(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.

[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]

(c) An expression for the period of the satellite’s orbit.

[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]

A satellite is a moon, planet or machine that orbits a planet or star. For example, Earth is a satellite because it orbits the sun

a) For this exercise we must use Newton's second law with the gravitational force

F = ma

[tex]ma =G\sqrt{\dfrac{mM_e}{(R+R_e)}[/tex]

the acceleration of the satellite is centripetal

[tex]a=\dfrac{v^2}{R+R_e}[/tex]

we substitute

[tex]G\dfrac{mM_e}{(R+R_e)}=m\dfrac{v^2}{(R+R_e)}[/tex]

[tex]G\dfrac{M_e}{(R+R_e)}=v^2[/tex]

[tex]v=\sqrt{G\dfrac{M_e}{(R+R_e)}[/tex]

b) mechanical energy is the sum of kinetic energy plus potential energy

        Em = K + U

[tex]Em =\dfrac{1}{2}m v^2 - \dfrac{G m M} {(R + R_e)}[/tex]

we substitute the expression for the velocity

[tex]E_m=\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}-G\dfrac{M_e}{(R+R_e)}[/tex]

[tex]E_m=-\dfrac{1}{2}G\dfrac{M_e}{(R+R_e)}[/tex]

c) as the orbit is circulating, the velocity modulus is constant

[tex]v=\dfrac{d}{t}[/tex]  

in a complete orbit the distance traveled of the circle is

[tex]d = 2\pi (R + R_e)[/tex]

where time is called period

[tex]v = 2\pi (R + R_e)[/tex]

[tex]T = \dfrac{2\pi (R + R_e)} { v}[/tex]

we substitute the speed value

[tex]T = 2\pi (R + R_e) . \sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]

[tex]T=2\pi\sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]

(a) An expression for the speed of the satellite in its orbit.

[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]

(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.

[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]

(c) An expression for the period of the satellite’s orbit.

[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]

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Answer:

Distance is a scalar quantity while displacement is a vector quantity

Explanation:

A scalar quantity represents only the magnitude and does not give any detail about the direction of the quantity for example distance. Distance can be any length measured in any direction (no specific direction)

However, a vector quality represents both the magnitude and direction. For instance displacement is a vector quantity. If direction is not defined then displacement becomes equal to distance.

The force due to gravity here is 14.7 N. The net force acting on the brick is 9.5 N and the acceleration of the brick is then 6.3 m/s².

Friction is a resistive force which opposes the motion of an object. The product of normal force by gravity and frictional coefficient gives the frictional force.

Given that, mass = 1.50 Kg

force by gravity = mg = 1.50 × 9.8 m/s² = 14.7 N

parallel component = 14.7 cos 34.4 = 12.12 N

perpendicular component  = 14.7  sin 34.4 = 8.30 N

The frictional force = mg × coefficient of friction

= 14.7 × 0.35 = 5.14 N

then net force = 14.7 - 5.14 = 9.5 N

Acceleration of the brick = net force/mass

a = 9.5 N/1.50 Kg = 6.30 m/s²

Hence, acceleration on the brick is 6.30 m/s².

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Answer:

800J

Explanation:

W = Fs, Work equals force times displacement

in this case, the force is 400N and the displacement is 2 meters.

The regular SI unit for work is joules

The mechanical advantage of the lever is 3.5.

The ideal mechanical advantage of the lever is 4

The efficiency of the lever is 87.5 %.

The mechanical advantage of the lever is the ratio of the load overcame by the lever to the effort applied by the lever.

M.A = Load / Effort

From the graph we will find the change in load to change in effort is calculated as;

M.A = ( 160 N - 20 N ) / ( 50 N - 10 N )

M.A = 3.5

The ideal mechanical advantage of the lever is calculated as follows;

I.M.A = distance travelled by effort / distance travelled by load

I.M.A = ( 2 m ) / ( 0. 5 m )

I.M.A = 4

The efficiency of the lever is calculated as follows;

E = M.A / I.M.A x 100%

E = ( 3.5 / 4 ) x 100%

E = 87.5 %

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The radius of the hydrogen-atom Bohr orbit shown in the figure is 5.3 nm.

The path that hypothetical electrons take around the nucleus is known as Bohr's orbit.

These orbits are described by Bohr in his hypothesis of the structure of an atom as energy levels or shells where electrons move in a fixed circle around the nucleus.

These orbits resemble solar system orbits, with the exception that they are attracted by electrical forces rather than gravity. The term "ground state" refers to the amount of energy that an electron typically occupies.

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Answer:

Is at the pivot of the wheel

Answer:

Explanation:

The mass of the ball can be found by using the formula

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

f is the force

a is the acceleration

We have

[tex]m = \frac{15}{5} = 3 \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

1.66 m/s

Explanation:

Work or kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex]

[tex]4864=\frac{1}{2} (3540)v^{2}[/tex]

v = 1.66 m/s

The linear expansion of the wrought iron is 0.0038 feet.

The linear expansion of the wrought iron is the increase in length of the wrought iron due to change or increase in temperature.

Mathematically, the formula for the linear expansion of metals is given as;

ΔL = αLΔθ

where;

The coefficient of thermal expansion of wrought iron = 0.0000067 / ⁰F

The linear expansion of the wrought iron is calculated as follows;

ΔL = αLΔθ

ΔL = ( 0.0000067/ ⁰F   x   8 ft    x   70 ⁰F )

ΔL = 0.0038 ft

Thus, the linear expansion of the wrought iron is a function of the increase in temperature of the iron and the original length of the iron.

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