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Ludmilka [50]
3 years ago
5

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?
Physics
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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Anit [1.1K]
 <span>(a) 

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get: 

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component. 

(b) 
To much the same end do we derive the vertical component: 

18.0*sin37.5 = v_y = 10.96 ms^-1 

Which we then divide by acceleration, a_y, to derive the time till maximal displacement, 

10.96/9.8 = 1.12 s 

Finally, doubling this value should yield the particle's total time with r_y > 0 

<span>2.24 s

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6 0
3 years ago
 Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc
slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

F=55\ kg\times 1.4\ m/s^2

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0
3 years ago
Two steamrollers begin 100 m apart and head toward each other, each at a constant speed of 1.00 m/s . At the same instant, a fly
Debora [2.8K]

Answer:

The fly travels 2.4 m

Explanation:

Since the Two steamrollers begin 100 m apart and head toward each other, each at a constant speed of 1.00 m/s, we can find the time until they crash by the formula:

Distance = Speed × Time

Time = Distance /Speed

Time = (100 m) / (1 m/s)

Time = 100 hours

Now, the fly will spend the same amount of time traveling as the steamrollers.

Since the fly moves at a speed of 2.4 m/s and we have a time of one hour the steamroller take to collide, then the fly will go a distance of;

Distance = speed x time = 2.4 × 1 = 2.4 m

4 0
3 years ago
Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical v
Brut [27]

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

l = 2a

The distance of the center of the square from each of the corners is

r = \frac{2a}{2}= a

The potential electric at the center due to each cornet charge is

V_1 = \frac{kQ_1}{r_1}

V_2 = \frac{kQ_2}{r_2}

V_3 = \frac{kQ_3}{r_3}

V_4 = \frac{kQ_4}{r_4}

The total electric potential at the center of the given square is

V = V_1+V_2+V_3+V_4

V = \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}+\frac{kQ_3}{r_3}+\frac{kQ_4}{r_4}

Al the charges are equal, and the distance are equal to a, then

V = \frac{kQ}{a}+ \frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a}

V = \frac{4kQ}{a}

Therefore the correct option is E.

3 0
3 years ago
The function x = (7.8 m) cos[(5πrad/s)t + π/3 rad] gives the simple harmonic motion of a body. At t = 4.4 s, what are the (a) di
VMariaS [17]

Answer:

a.3.84m

b.-106.67m/s

c.947.3m/s^2

d.70.17 rad

e.2.5Hz

d.0.4secs

Explanation:

Given x=(7.8)cos[5πrad/s)t+π/3)]

a.Displacement at t=4.4

7.8cos(5π*4.4+π\3)=3.84m

b.velocity

V= dx/dr=-5π(7.8)sin(5πrad/s)t+π\3

at t=4.4

-5π(7.8)sin(5π*4.4+π\3)=-106.67m/s

c.acceleration

a=d^2x/dr^2

-(5π)^2(7.8) cos (5π*t+π\3)

at t=4.4

-(5π)^2(7.8)cos(5π*4.4+π\3)=-947.3m/s^2

d. Phase =(5πrad/s)t+π\3

At t=4.4

5π×4.4+π\3=70.17 rad

e.frequency

Given x= 7.8cos(5πt+π\3

Compare with x=Acos(2πft)

2πft=5πt

F=2.5Hz

f.T=1\f

T=1/2.5=0.4sec

6 0
3 years ago
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