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Delvig [45]
3 years ago
14

An object has 90,000 J of kinetic energy and is moving at 12 m/s. What is the objects mass?

Physics
1 answer:
trapecia [35]3 years ago
4 0

Answer:

plug it into your calculator

Explanation:

Kinetic energy = KE = Joules = J

KE= 1/2mv^2

90,000 = 1/2 m (12^2)

180,000 = m (144)

m = 180,000 / 144

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The average coefficient of linear expansion of copper is 1.7 10-5 (°c)−1. the statue of liberty is 93 m tall on a summer morning
sammy [17]

Let the rise in temperature be 5^0C

The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the  coefficient of linear expansion, Δt is the change in temperature.

Here l = 93 m, α = 1.7*10^{-5}  ^0C^{-1}, and Δt = 5^0C

So expansion in length = 93*1.7*10^{-5}*5 = 0.007905 m = 0.79*10^{-3}m

So order of magnitude in change in length = -3


3 0
3 years ago
Read 2 more answers
What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub
aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse (  F \times t). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

s=ut+\frac{1}{2} gt^2

Here, u is initial velocity which is zero.

s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }.

Thus, impulse

= F \times \sqrt{\frac{2s}{g} }

From Newton`s second law,

F =mg

Therefore, impulse

= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}

Given,  m = 7.3 kg and s = 2.0 m

Substituting these values, we get

Change in momentum = impulse  

= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns.

8 0
3 years ago
A sealed container filled with gas is heated. What happens?
Luda [366]

The internal pressure increases as the gas is heated

5 0
3 years ago
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Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?​
Sliva [168]

Answer:

\huge\star{\underline{\mathtt{\blue{Answer}}}}\huge\star...

<h2>PE=<em>work done</em></h2><h2><em>m</em><em>gh</em><em>=</em><em>2</em><em>0</em><em>×</em><em>1</em><em>0</em><em>×</em><em>2</em><em>0</em><em>.</em><em>.</em></h2>

\huge\boxed{\fcolorbox{white}{blue}{mgh=4000}}

.

<em>I </em><em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

7 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
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