Physics, 12.01.2020 14:31 bryanmcmillianjr

Ablock of mass m = 2.50 kg is pushed d = 2.10 m along a frictionless horizontal table by a constant applied force of magnitude f = 20.0 n directed at an angle θ = 25.0° below the horizontal as shown in the figure below. a block labeled m is on a horizontal surface. an arrow labeled vector f points downward and to the right at an angle θ above the horizontal, and acts upon the upper left corner of the block. a faded image of the block is a distance d to the right of the block.
(a) determine the work done by the applied force. j
(b) determine the work done by the normal force exerted by the table. j
(c) determine the work done by the force of gravity. j
(d) determine the work done by the net force on the block. n

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Answer from: trinityscute81

(a) 38.1 J

The work done by the applied force is given by:

$W=F d cos \theta$

where

F = 20.0 N is the magnitude of the applied force

d = 2.10 m is the displacement of the block

$\theta=25.0^{\circ}$ is the angle between the direction of the force and the displacement of the block

Substituting into the formula, we find

$W=(20.0 N)(2.10 m)(cos 25.0^{\circ})=38.1 J$

(b) 0

As before, the work done by the normal force exerted by the table is:

The work done by the applied force is given by:

$W=F d cos \theta$

where

F is the magnitude of the normal force

d = 2.10 m is the displacement of the block

$\theta=90.0^{\circ}$ is the angle between the direction of the force and the displacement of the block, and it is 90 degrees because the block is moving along a horizontal surface, while the normal force is vertical

Since $cos 90^{\circ}=0$ , therefore, the work done by the normal force is zero as well.

As in part c), the force (gravity) is perpendicular to the displacement (because gravity acts vertically, while the displacement of the block is horizontal), therefore the work done by gravity is zero as well.

(d) 38.1 J

Since the surface is frictionless, there is no friction acting on the block. Therefore, there are only 3 forces acting on the block: the applied force F, the gravity, and the normal force exerted by the table. Since the last 2 forces do zero work on the block, the total work done on the block is just the work done by the applied force, therefore 38.1 J.

Answer from: Quest

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

Answer from: Quest

c because the news are a source of information of things that happen on the daily and shows evidence about that particular topic

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