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shutvik [7]
3 years ago
8

Find the energy in joules required to lift a 55.0 megagram object a distance of 500cm

Physics
1 answer:
pantera1 [17]3 years ago
5 0

1,000 grams = 1 kilogram
so 55 megagrams = 55,000 kilograms

100 cm = 1 meter
so 500 cm = 5 meters

Acceleration of gravity on Earth = 9.8 m/s²

Weight = (mass) x (gravity)

========================================

Work = increase in potential energy =

               (weight) x (height) =

             (mass) x (gravity) x (height) =

             (55,000 kg) x (9.8 m/s²) x (5 m) =

                     2,695,000 joules .

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A satellite orbits the earth at a speed of 8.0x 10^3m/s . Calculate the period of orbit of the satellite if the distance between
gladu [14]

Hi there!

The period of an orbit can be found by:

T = \frac{2\pi r}{v}

T = Period (? s)
r = radius of orbit (6400000 m)

v = speed of the satellite (8000 m/s)

This is the same as the distance = vt equation. The total distance traveled by the satellite is the circumference of its circular orbit.

Let's plug in what we know and solve.

T = \frac{2\pi (6400000)}{8000} = \boxed{5026.55 s}

8 0
1 year ago
An object falling straight down without air resistance is said to be exhibiting
Tanzania [10]

Answer:

Explanation:

If the object is falling straight down it is in free fall.  The difference between that and two-dimensional motion is that 2D motion is parabolic (projectile)

6 0
3 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
loris [4]

Answer:

The acceleration of the crate is 1.82\ m/s^2.

Explanation:

Given that,

Force, F = 750 N

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The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

F=ma\\\\F-f=ma

f is frictional force, f=\mu N=\mu mg

F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2

So, the acceleration of the crate is 1.82\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
A spotlight on the ground shines on a wall 12 meters away. If a man 2 meters high walks away from the spotlight towards the wall
nordsb [41]

Answer: The length of the shadow on the wall is decreasing by 0.6m/s

Explanation:

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At that moment, his shadow on the wall is y=BC.

The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:

ADAB=DEBC

8/12=2/y,∴y=3 meters

If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.

(12−x) /12=2/y

1− (1 /12x )=2 × 1/y

Let's take derivatives of both sides:

−1 / 12dx = −2 × 1 / y^2 dy

Let's divide both sides by dt:

−1/12⋅dx/dt=−2/y^2⋅dy/dt

At the specified moment:

dxdt=1.6 m/s

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Let's plug them in:

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5 0
3 years ago
Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the video, the mass of t
soldi70 [24.7K]

Answer:

a) three times greater

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Since water increases 20 ºC and the metal 60 ºC, it was neccesary 3 times of the quantity of heat to the same amount of mass of the two different susbtances.

Hope it was useful

5 0
3 years ago
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