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deff fn [24]
3 years ago
10

Sir Dudley the Dextrous is loading cannons on top of the castle's tallest towers in preparation for an expected attack. He slips

, and a cannonball of mass 14.0 kg 14.0 kg rolls to the edge of the tower, which has a perfectly horizontal surface, and falls to the ground below. It takes the cannonball 3.15 s 3.15 s to fall to the ground, and air resistance is negligible. How much work is done by gravity on the falling cannonball?
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

6667.9 J

Explanation:

We are given that

Mass of cannon.m=14 kg

Time,t=3.15 s

Initial velocity,u=0

We have to find the work done by gravity on the falling cannon ball.

g=9.8 m/s^2

h=ut+\frac{1}{2}gt^2

Substitute the values

h=0+\frac{1}{2}\times 9.8\times (3.15)^2=48.6 m

Work done,W=mgh

Using the formula

W=14\times 9.8\times 48.6=6667.9 J

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A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that
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Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

Find the Speed of the Proton

The formula for Kinectic Energy =

K.E = 1/ 2 mv²

Where

m = mass of the proton

v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

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Step 3

Find the Magnetic Field of that region of space

The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

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