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N76 [4]
3 years ago
5

A small object is sliding along a level surface with negligible friction, and has mass m, and constant speed v_0, when it collid

es with a brick resting on the surface. If the object recoils backward at (v_0)/3, and is in contact with the brick for time delta t, what average force did the brick exert on the object? Assume the object is moving along x-direction.
A. F_net = 0
B. F_net= -((mv_0)/delta t)
C. F_net = -((4mv)/(3delta t))
D. F_net = -((mv_0)/ delta t)
E. F_net = -((3mv_0)/(2delta t))
F. F_net = -((2mv_0)/(3delta t))
Physics
1 answer:
VMariaS [17]3 years ago
7 0

To solve this problem we will apply the concepts related to the Impulse-Momentum theorem. On this theorem it is specified that the Momentum is defined as the product between the mass and the velocity of the object, and the Impulse as the product between the Force and time. Both concepts fully comparable to each other. Mathematically they can be described as,

\Delta p = m \Delta v

And

\Delta P = F_{net} \Delta t

Here,

m = mass\\\Delta v = \text{Change in velocity}\\F_{net} = \text{ Net Force}\\\Delta t = \text{Change in time}\\

According to the value given we have that the change in velocity in the momentum is

\Delta p = m(-v_0 - \frac{v_0}{3})

\Delta p = -4 m \frac{(v_0)}{3}

Using this value to find the Net force we have that

F_{net} = \frac{\Delta P}{\Delta t}

F_{net} =\frac{-4 m \frac{v_0}{3}}{\Delta t}

F_{net}  = \frac{-4 mv_0}{3 \Delta t}

Therefore the correct answer is C.

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4. How does a molecule differ from an atom
tia_tia [17]

Answer:

Explanation:

In a molecule, atoms are bonded together by single, double, or triple bonds. An atom has a nucleus surrounded by electrons. ... So another difference between atoms and molecules is that when similar atoms combine together in varying numbers, molecules of different properties can be formed.

8 0
3 years ago
Read 2 more answers
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
choli [55]

Answer:

They can be seen from a distance of 4.372 kilometers.

Explanation:

Using the Reyligh creterion for diffraction through a circular aperture we have

\frac{x}{D}=\frac{1.22\lambda }{d}

where symbol's have their usual meaning

thus applying values we get

D=\frac{dx}{1.22\lambda }

\therefore D=\frac{0.633\times 4.61\times 10^{-3}}{1.22\times 547\times 10^{-9}}\\\\D=4372.77m\\=4.372km

5 0
3 years ago
Calculate Vector component in Y if the hypotenuse is 32 and angle is 45
Lerok [7]

Answer:

The correct option is;

c. 22.6

Explanation:

The given parameters are;

The hypotenuse of the vector = 32

The angle of the vector = 45°

Therefore, the vector component in the y-axis is given as follows;

v_y = v \times sin(\theta)

Substituting the values from the question gives;

v_y = 32 \times sin(45^{\circ}) \approx 22.6

The vector component in the y-axis, v_y, is approximately 22.6.

8 0
2 years ago
A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is
Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_{f} = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_{f}

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

     

initial energy

         K₀ = ½ m v²

final energy

        K_{f} = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

         

the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

5 0
3 years ago
1000 kg of water initially at 6 m/s runs through a hydro-generator. If the water leaves the generator at velocity of 4 m/s, and
r-ruslan [8.4K]

Answer:

8.00 kJ

Explanation:

The first thing is to determine what quantities are there.

the mass of water = 1 000 kg

initial velocity, u = 6 m/s

final velocity, v = 4 m/s

the generator is operating at 100 % efficiency, so there is no energy loss.

The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.

The kinetic energy is calculated as follows:

Ek = 1/2 mv²

    = 1/2×(1 000)× (4)²

    = 8 000 J/s

    = 8.00 kJ  Ans

4 0
3 years ago
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