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erastova [34]
2 years ago
10

Identical cannon balls are fired with the same force, one each from four cannons having respective bore lengths of 1.0 meter, 2.

0 meters, 2.5 meters and 3.0 meters. Which cannon will have the minimum impulse on the ball if the force acting on the ball increases with the length of the bore?
A.) 3.0 meters

B.) 1.0 meters

C.) 2.5 meters

D.) 2.0 meters
Physics
1 answer:
Gala2k [10]2 years ago
4 0

To answer this question, we must bear in mind the following considerations that are mentioned in the statement:

The cannon balls are identical and shoot with the same force

The force acting on the cannonball increases with the length of the hole.

You want to know which cannon will have the least momentum on the ball.

Then, the force on the ball increases as the barrel length increases and the impulse depends on the magnitude of the force, then, the cannon that will have the minimum impulse will be the 1 meter one.

The answer is option B.


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Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of
zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, m=40\ kg

Radius of the barrel is, R=2.90\ m

Number of revolutions are, n =10

Time taken for 10 revolutions is, t=29.3\ s

Therefore, the time period of the child is given as:

T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s

Now, angular velocity is related to time period as:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s

Now, centripetal force acting on the child is given as:

F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0
2 years ago
Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
2 years ago
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