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Lapatulllka [165]
3 years ago
7

A 48 kg cart is sitting motionless at the top of the hill. At what height, did the cart start at, to reach 88435 J of energy at

the end of the ride? Group of answer choices 19 meters 456 meters 188 meters 0.19 meters
Physics
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer : The height is 188 meters  

Explanation :   When the cart reached at the end from top of hill then the cart have potential energy .

Given that,

Potential energy = 88435 J

Mass of cart = 48 kg

We know that,

The potential energy is

mgh =88435

h= \dfrac{88435}{48\times9.8}

h = 187.9 = 188\ meters

So, the height of the top is 188 meters.

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1 year ago
Where do seasons alternate between a very dry season and a monsoon season, with a stable warm climate?
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7 0
3 years ago
If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat
Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

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3 0
3 years ago
I need help with these. Please show workings<br>​
Sauron [17]

Answer:

Imp = 25 [kg*m/s]

v₂= 20 [m/s]

Explanation:

In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.

1)

(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})

where:

m₁ = mass of the object = 5 [kg]

v₁ = initial velocity = 0 (initially at rest)

F = force = 5 [N]

t = time = 5 [s]

v₂ = velocity after the momentum [m/s]

(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]

2)

(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]

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2 years ago
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