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2 years ago

What height would cause an object to have no gravitational potential energy

1 answer:

OverLord2011 [107]2 years ago

5 0

The height is zero.
That means whatever place is your reference for height measurement,
the object is lying right there at that height.

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10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that

Oliga [24]

Answer:

Explanation:

10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that her spaceship's length is 1150 m. (c = 3.00 × 108 m/s) (a) How fast is the rocket moving relative to Earth? (b) What is the TOTAL energy of a 75.0-kg crewman as measured by (i) the captain in the rocket and (ii) the Earthling?

3 0

2 years ago

Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, re

ch4aika [34]

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of $62 lb / ft ^ 3$ (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

$W = \gamma A * \int_0^15 dy$

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

$W = (62)(14*7)\int^{15}_0 (15-y)dy$

$W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0$

$W = (14*7*62)[15(15)-\frac{(15)^2}{2}]$

$W = 683550ft-lbs$

Therefore the total work in the system is $683550ft-lbs$

6 0

3 years ago

An object starts at 4m/s and accelerates to 6m/s in 10 seconds. Find its displacement.

Natalka [10]

Answer:

Explanation:

The 2 equations we need here are, first:

$a=\frac{v_f-v_0}{t}$ and then once we solve for the acceleration here:

$v^2=v_0^2+2a$ Δx

Solving for acceleration:

$a=\frac{6-4}{10}=\frac{1}{5}\frac{m}{s^2}$ and now we will use that in the other equation:

$6^2=4^2+2(\frac{1}{5})$ Δx and

36 = 16 + $\frac{2}{5}$ Δx and

20 = $\frac{2}{5}$ Δx and

$\frac{5}{2}(20)=$ Δx so

Δx = 50 m

6 0

2 years ago

A 1.4 kg block slides across a rough surface such that it slows down with an acceleration of 1.25 m/s(squared). What is the coef

tino4ka555 [31]

F = ma

where 

F = frictional force 
m = mass of the block = 1.4 kg (given) 
a = acceleration of the block = 1.25 m/sec^2 (given) 

Substituting values, 

F = (1.4)(1.25) 

F = 1.75 N 

By definition, 

F = (mu)(Normal force) 

where 

mu = coefficient of friction 
Normal force = mg = 1.4*9.8 = 13.72 

Again, substituting appropriate values, 

1.75 = mu(13.72) 

mu = 0.128

5 0

2 years ago

Single Stage 0.0/3.0 points (graded) Consider a rocket carrying 100,000 kg of propellant, 10,000 kg of structure, and 5000 kg of

Nonamiya [84]

Answer: The Propellant fraction is 0.87.

The payload fraction is 0.04.

Δv = 8991.81 m/s

Explanation: To determine the fractions, first, calculate the total mass of the rocket:

$m_{t} = m_{prop} + m_{str} + m_{pay}$