In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after are given. The collision is (Show your work-no work shown = ZERO POINTS) 1.8 m/s 0.2 m/s 0.6 m/s 1.4 m/s 4 kg 6 kg 4 kg 6 kg Before After A) perfectly elastic. B) partially inelastic. C) completely inelastic. D) characterized by an increase in kinetic energy E) not possible because momentum is not conserved.

Answers

Answer 1

Answer:

When two bodies come into close touch with one another, a collision occurs. In this instance, the two bodies quickly exert forces on one another. The collision changes the energy and momentum of the bodies that are interacting.

Briefing

the system's initial kinetic energy, KEi, is equal to 0.5 * 4 * 1.8 2 plus 0.5 * 6 * 0.2 2 J.

KEi = 6.6 J

The system's ultimate kinetic energy, KEf

, following the collision is equal to 0.5 * 4 * 0.6 + 0.5 * 6 * 1.4 J.

KEf = 6.6 J

since KEi = KEf

Perfectly elastic is the collision

the appropriate response is A) completely elastic.

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A certain x-ray tube requires a current of 7 mA at a voltage of 80 kV. The rate of energy dissipation is:

Answers

Answer:

560 watts

The rate of energy dissipation is 560 W

Explanation:

Rate of energy dissipation is the rate of energy consumption in the x-ray tube.

P = VI

Given;

Voltage V = 80 kV = 80,000

Current I = 7mA = 0.007 A

Substituting the given values;

P = 80,000 V × 0.007A

P = 560 Joules per second

P = 560 watts

The rate of energy dissipation is 560 W

If 80 joules of work were necessary to move a 5 newton box, how far was the box moved?

Answers

Explanation:

work=force/distance

work=80

force=5

putting value of force and work we get

80=5/distance

5/80=distance

1/16=distance

0.0625m

6.25cm

A proton and an alpha particle (helium nucleus consisting of two protons and two neutrons) are accelerated from rest across the same potential difference. Assume the proton mass and the neutron mass are roughly the same and neglect any relativistic effect. Compared to the final speed of the proton, the final speed of the alpha particle is?1. less by a factor of 22. less by a factor of √ 23. less by a factor of 44. greater by a factor of 25. the same

Answers

Answer:

option B

Explanation:

we know,

change in energy is equal to

$W = (1)/(2)m(v^2 - u^2)$

$W = (1)/(2)m(v^2 - 0^2)$

$W = (1)/(2)m v^2$

$q = (1)/(2)m v^2$

proton mass and the neutron mass are roughly the same

so,

$q \alpha m v^2$

now,

$(q_p)/(q_(\alpha)) = (m_p v_p^2)/(m_(\alpha)v_(\alpha)^2)$

$(q_p)/(q_(\alpha)) = (m_p v_p^2)/(2 m_pv_(\alpha)^2)$

we know,

mass of alpha particle is four times mass of the mass of proton.

mα = 4 m_p

$(e)/(2e) = ( v_p^2)/(4 v_(\alpha)^2)$

$( v_p^2)/(v_(\alpha)^2) = 2$

$v_(\alpha)^2 =( v_p^2)/(2)$

$v_(\alpha)=( v_p)/(√(2))$

less by a factor of √2

Hence, the correct answer is option B

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

Plzzz helppppp!!! I need answers A, B, C & D

Answers

Answer: the answer i for c is yes 0& 10

Explanation:

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

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