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Vika [28.1K]
4 years ago
12

A fireman is sliding down a fire pole. As he speeds up, he tightens his grip on the pole, thus increasing the vertical frictiona

l force that the pole exerts on the fireman. When the force on his hands equals his weight, what happens to the fireman?
A. The fireman comes to a stop.
B. The fireman descends with slower and slower speed.
C. The fireman descends with a smaller but non-zero acceleration.
D. The fireman continues to descend, but with constant speed.
E. The acceleration of the fireman is now upward.
Physics
1 answer:
Gnesinka [82]4 years ago
6 0

Answer:

The fireman continues to descend but with constant speed.

Explanation:  

According to Newton first law of motion, The object which is rest continues to remain at rest or object moving with constant speed move with constant velocity when acted upon by an unbalanced force.

When the net force acting on the object becomes zero the object moves with constant speed.

Hence when the frictional force acting on the fireman hands equals to his body weight , the net force acting on the fireman becomes zero and thus fireman continues to descend with constant speed.

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Andy has two strong magnetic marbles of equal size. What will happen when he puts the two marbles about 2 centimeters apart?
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I believe it would be C
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4 years ago
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Explain imprinting in relation to Lorenz's geese. What would most likely have happened if the first thing the goslings saw was a
iragen [17]

If the first thing that the goslings saw was a dog, they would have followed the dog as a mother.

Imprinting refers to the process of training an animal to bond with anything it sees after birth even if it is not its real mother. Lorenz first achieved imprinting in 1935 using geese which followed him as their mother shortly after they were born.

If the geese were exposed to a dog, they could also have seen the dog as their mother and followed it accordingly shortly after birth.

Learn more: brainly.com/question/11401513

7 0
2 years ago
Trina makes a diagram to summarize how to use the right-hand rule to determine the direction of the magnetic force on each type
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Labels that belong in regions X, Y, Z;

B) X: Back of hand, for force

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7 0
3 years ago
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Dos cargas puntuales (q1 y q2 ) se atraen inicialmente entre sí con una fuerza de 550 N , si la separación entre ellas es de 90
antiseptic1488 [7]

Answer:

q2 = 9.02*10^{-4}C

Explanation:

To find the value of the other charge you use the Coulomb's law:

F=k\frac{q_1q_2}{r^2}

k: Coulomb's constant = 8.98*10^{9}Nm^2/C^2

q1: charge 1 = 55*10^{-6}C

r: distance between charges = 90cm = 0.9m

F: electric force = 550N

By doing q2 the subject of the formula and replacing you obtain:

q_2=\frac{r^2F}{kq_1}=\frac{(0.9m)^2(550N)}{(8.98*10^9Nm^2/C^2)(55*10^{-6}C)}=9.02*10^{-4}C

hence, the value of the other charge q2 is 9.02*10^{-4}C

6 0
3 years ago
A(n) 14 g bullet is fired into a(n) 121 g block of wood at rest on a horizontal surface and stays inside. After impact, the bloc
kotegsom [21]

Answer:

33.14 m/s

Explanation:

The mass of the block is 121g or .121 kg. As the bullet is lodged in the block the total mass is 121+14 = 135 g or 0.135 kg.

The frictional force that makes the block come to a stop is normal force* coefficient of friction = 0.135 * 9.8 * 0.7 = 0.9261 N

As the block comes to rest after sliding for 8.3 meters the energy it was given by the bullet is

0.135 * 9.8 * 0.7 * 8.3

= 7.69 Nm

Now this energy is provided the bullet. So the energy in the bullet was equal to

1/2 * mv² = 0.5 * 14 * v².

0.5 * 0.014 * v^2 = 0.135 * 9.8 * 0.7 * 8.3 = 7.69

=> 0.007 * v² = 7.69

=> v² = 7.69 / 0.007

=> v² = 1098.57

=> v = √1098.57

=> v = 33.14 m/s

8 0
3 years ago
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