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3 years ago

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

he proton must impact the nucleus with a kinetic energy of 1.80 MeV. Assume the nucleus remains at rest.
Required:
a. With what speed must the proton be fired toward the target?
b. Through what potential difference must the proton be accelerated from rest to acquire this speed?

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

Explanation:

kinetic energy required = 1.80 MeV

= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 2.88 x 10⁻¹³ J

If v be the velocity of proton

1/2 x mass of proton x v² = 2.88 x 10⁻¹³

= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

v² = 3.45 x 10¹⁴

v = 1.86 x 10⁷ m /s

If V be the potential difference required

V x e = kinetic energy . where e is charge on proton .

V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³

V = 1.8 x 10⁶ volt .

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In which of the following cases will the normal force on a box be the greatest? A. When the box is placed in a stationary elevat

Butoxors [25]

Answer:

D. When the box is placed in an elevator accelerating upward

Explanation:

Looking at the answer choices, we know that we want to find out how the normal force varies with the motion of the box. In all cases listed in the answer choices, there are two forces acting on the box: the normal force and the force of gravity. These two act in opposite directions: the normal force, N, in the upward direction and gravity, mg, in the downward direction. Taking the upward direction to be positive, we can express the net force on the box as N - mg.

From Newton's Second Law, this is also equal to ma, where a is the acceleration of the box (again with the upward direction being positive). For answer choices (A) and (B), the net acceleration of the box is zero, so N = mg. We can see how the acceleration of the elevator (and, hence, of the box) affects the normal force. The larger the acceleration (in the positive, i.e., upward, direction), the larger the normal force is to preserve the equality: N - mg = ma, N = ma+ mg. Answer choice (D), in which the elevator is accelerating upward, results in the greatest normal force, since in that case the magnitude of the normal force is greater than gravity by the amount ma.

4 0

3 years ago

Read 2 more answers

A 10 kg frictionless cart is pushed at a constant force of 5.0 N for a distance of 10 m. The work done on the cart is 50J.

kykrilka [37]

1) Final kinetic energy of the cart: B) 50 J

2) Final speed of the cart: C) 3.2 m/s

3) Height reached along the ramp: A) 0.5 m

Explanation:

1)

We can solve this part of the problem by using the work-energy theorem, which states that the work done on an object is equal to the kinetic energy gained by the object itself. Mathematically:

$W=K_f - K_i$

where

W is the work done

$K_f$ is the final kinetic energy

$K_i$ is the initial kinetic energy

In this problem, the work done on the cart is

W = 50 J

And assuming it starts from rest, its initial kinetic energy is zero:

$K_i = 0$

Therefore, the final kinetic energy is:

$K_f = K_i + W=0+50=50 J$

2)

The kinetic energy of an object is the energy possessed by an object due to its motion; it is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the cart in this problem, we have:

K = 50 J is its final kinetic energy

m = 10 kg is the cart

Therefore, solving the formula for v, we find its speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(50)}{10}}=3.2 m/s$

3)

We can think this problem in terms of conservation of energy. In fact, as the cart rolls up the ramp, its kinetic energy is converted into gravitational potential energy, which is given by

$PE=mgh$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the heigth of the cart

When the cart reaches the maximum height, all the kinetic energy has been converted into potential energy, so we can write:

$K=PE\\\frac{1}{2}mv^2=mgh$

Re-arranging,

$h=\frac{v^2}{2g}$

And since we know the initial speed of the cart along the ramp,

v = 3.2 m/s

we can find the maximum height reached along the ramp:

$h=\frac{3.2^2}{2(9.8)}=0.5 m$

Learn more about work, kinetic energy and potential energy:

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3 years ago

You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc

pickupchik [31]

a) $15.4^{\circ}$

b) 5.2 m/s

c) 151.2 N

Explanation:

a)

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

$mg sin \theta$

where

m =16 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

$\theta$ is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

$F=ma\\mgsin \theta = ma$

where a is the acceleration.

From the equation above we get

$a=g sin \theta$

And we are told that the acceleration must not exceed

$a=2.6 m/s^2$

Substituting this value and solving for $\theta$ , we find the maximum angle of the ramp:

$\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}$

b)

Here we are told that the vertical distance of the ramp is

$h=1.4 m$

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

$GPE=KE\\mgh=\frac{1}{2}mv^2$

where:

m = 16 kg is the mass of the box

$g=9.8 m/s^2$

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s$

c)

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

$mg cos \theta$

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

$N-mg cos \theta$

and by substituting:

m = 16 kg

$g=9.8 m/s^2$

$\theta=15.4^{\circ}$

We can find the normal force:

$N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N$

8 0

3 years ago

Simple physics question, check the document. Should take about 3-5 minutes.

Ahat [919]

Answer:

The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N

Explanation:

Forces on block 4.3 kg are:

63N to the right and R21 (contact force from the 6.3 kg block) to the left

Net force on 4.3 kg block is: 63 N - R21

Forces on the 6.3 kg block are:

R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.

So net force on the 6.3 kg block is: R12 - 11 N

According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").

Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:

a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2

solve for R by cross multiplication

6.3 (63 - R) = 4.3 (R - 11)

396.9 - 6.3 R = 4.3 R - 47.3

369.9 + 47.3 = 10.6 R

444.2 = 10.6 R

R = 444.2 / 10.6

R = 41.90 N

5 0

3 years ago

If the net force acting on a moving object CAUSES NO CHANGE IN ITS VELOCITY, what happens to the object's momentum?

SOVA2 [1]

If the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

<h3>What is momentum?</h3>

Momentum of a body in motion refers to the tendency of a body to maintain its inertial motion.

The momentum is the product of its mass and velocity.

This suggests that if the net force acting on a moving object causes no change in its velocity, the momentum of the object will remain the same.

Therefore, if the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

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