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marta [7]
3 years ago
14

A 30 N force toward the west is applied to an object. The object moves 50 m east during the time the force is applied. What is t

he change in kinetic energy of the object?
a) 1.0 J
b) 750 J
c) 1.7 J
d) -1500 J
Physics
1 answer:
katovenus [111]3 years ago
8 0

Answer:

D.<u>-1500Joules</u>

Explanation:

The change in kinetic energy of the object s equivalent to the workdone by the body in the west direction (negative x direction)

Workdone = Force * Distance

Given

Force = 30N

Distance moved by the object = 30m

Required

Kinetic energy

Kinetic energy = 30 * 50

Kinetic energy = 1500Joules

Since the body moves in the negative  direction, hence the kinetic energy will be <u>-1500Joules</u>

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_____ of rough surfaces reduces friction
Radda [10]

Answer:

Hey shaikaadil700 !

<u> </u><u>Lubricating</u><u> </u> of rough surfaces reduces friction.

Explanation:

• Lubricating is the smoothening or polishing of the surfaces

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3 0
3 years ago
Prob. 3: Manifestation of quantum phenomena (total 25 points) (a)-(e) 5 points each.
TEA [102]

Answer:

3A. This phenomenon can be seen in the discrete emission of the molecules.

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams

3D. This is due to the stimulated emission

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus)

Explanation:

This problem asks for some experimental explanations of various quantum phenomena.

3A. This phenomenon can be seen in the discrete emission of the molecules.

In the classical explanation all states or energies are allowed, therefore when emitting energy (photons) there should be a continuum, this is not observed

In the correct quantum explanation only some states are allowed, therefore the emission must be discrete, which is observed in the emission or absorption of molecules and atoms

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

The incorrect classical explanation that if the minimum energy was zero the electrons cannot rotate around the nuclei and the atom collapses, this does not happen

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams, which is evidence of the existence of two discrete states that we call spin, remember that a free electron beam has zero angular momentum.

3D. This is due to the stimulated emission that occurs when a photon passes through the emission zone, causing the atoms to have transitions and these emitted photons have the same initial photon location, the laser beam all photons are in phase and therefore it is coherent .

This is widely used for holographic and interference work

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus) leaves the atomic nucleus penetrating the barrier since its energy is lower than the nuclear barrier potential.

6 0
3 years ago
4. What is the instantaneous acceleration at t= 10 s?
vodka [1.7K]

Answer:

I am fairly certain the answer is 2m/s^2

Explanation:

6 0
3 years ago
A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta
creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

F_c=F_e=ma_c      (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

F_e=k\frac{qq'}{r^2}             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

a_c=\frac{v^2}{r}                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}

Finally, you replace the values of all parameters in the previous expression:

r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m

The radius of the circular trajectory is 0.22m

5 0
3 years ago
A 1.0 × 10^3 kg sports car is initially traveling at 15 m/s. The driver then applies the brakes for several seconds so that -25
algol13

Answer:

Ea = 112500[J]

Eb = 87500[J]

Explanation:

To solve this problem we must use the principle of energy conservation which tells us that the energy of a body plus the work done or applied by the body equals the final energy of a body.

This can be easily visualized by the following equation:

E_{A}+E_{friction}=E_{B}

Now we must define the energies at points A & B.

<u>For point A</u>

At point A we only have kinetic energy since it moves at 15 [m/s]

So the kinetic energy

E_{A}=\frac{1}{2}*m*v_{A}^{2}  \\E_{A}=\frac{1}{2} *1000*(15)^{2} \\E_{A}=112500[J]

The final kinetic energy can be calculated as follows:

112500-25000=E_{B}\\E_{B}=87500[J]

8 0
3 years ago
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