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2 years ago

What is the physics definition of work? How is work calculated? How would you calculate the amount of work done in a slapshot.

Physics

1 answer:

loris [4]2 years ago

3 0

Answer:

1) Work is a measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement. 2) Work is calculated through joules. 3) It's possible to calculate work by multiplying the amount of force applied by the distance that the force is applied. If we know the amount of force the stick applies to the puck and the distance that the stick is applying the force to the puck, we can figure out the amount of work done.

Explanation:

Hope this helped!- Nya~ :3

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You throw a rock horizontally out of a 7th story window. You time that it takes 3.7 seconds to hit the ground, and measure that

Nady [450]

Since we are only looking at the vertical height, we can use the free fall equation to find the height:
h = 0.5*g*t^2, where h is height in m, g is acceleration due to gravity (9.81 m/s^2), and t is time in seconds
h = 0.5*(9.81 m/s^2)*(3.7 s)^2
h = 67.15 m
Therefore, the 7th floor window is 67.15 m above ground level.

5 0

3 years ago

Define thermal conductivity.

ladessa [460]

Answer:

A measure of the ability of a material to transfer heat.

Explanation:

Please mark me as brainliest please

4 0

3 years ago

You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

Pavel [41]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

$s_1 = v_1t = 1.6t$

Similarly the distance (m) from your friend building to the restaurant:

$s_2 = v_2t = 1.2t$

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

$s_1 + s_2 = 5b$

$1.6t + 1.2t = 5b$

$2.8t = 5b$

$t = 5b/2.8 = 25b/14$

$s_2 = 1.2t = 1.2(25b/14) = 2.143b$

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

7 0

3 years ago

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang

adell [148]

Answer:

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work(friction) = 453.5J

work(gravity) = -453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work(friction) = mgsinθ .displacement

= (14) (9.81) (5.5sin36.9°)

= 453.5J

work(gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

work(gravity) = (14) (9.81) (5.5cos126.9°)

= -453.5J

8 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$