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3 years ago

9

Plzz help me please

1 answer:

olasank [31]3 years ago

6 0

Permanent magnet. An induced magnet would be created when a piece of iron (for example) is in contact with a magnet. Temporary magnets would be something like an electromagnet. Bar magnets are permanently magnetic unless we heat them or hammer them to cause their domains to loose alignment.

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everything

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What is the mass of a football player if they are running at 3.6 m/s with a momentum of 62.5kgm/s?

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momentum = mass * velocity

62.5 = mass * 3.6

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4 years ago

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The law of conservation of energy states that (4 points)

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3 years ago

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc

____ [38]

Answer:

Part a)

$a = 0.36 m/s^2$

Part b)

$a = -1.29 m/s^2$

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

$F_x = Fcos25$

$F_x = 80 cos25 = 72.5 N$

$F_y = Fsin25$

$F_y = 80 sin25 = 33.8 N$

now in vertical direction we can use force balance for the box to find the normal force on it

$F_n + F_y = mg$

$F_n = (25)(9.81) - 33.8$

$F_n = 211.45 N$

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

$F_k = (0.300)(211.45) = 63.44 N$

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

$f_k = (0.300)(207.7) = 62.3 N$

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

7 0

4 years ago