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Roman55 [17]
3 years ago
15

A monkey with 4.5 kg of mass falls while swinging from a tree 10 meters above the ground. What is its GPE while it is on the tre

e? SHOW WORK​
Physics
1 answer:
Tpy6a [65]3 years ago
7 0

Answer:

no se

jv,,,,,hgbbbbbbbbbbbbbbbbbbbbbbbbdxj hryjjccccccccccccccccccccccccccccccccc

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1. A sample consisting of 1.0 mol CaCO3 (s) was heated to 800oC when it is decomposed. The heating was carried out in a containe
ExtremeBDS [4]
Expansion work against constant external pressure: w=-pex Δ Δ V 3. The attempt at a solution . I tried following that. Because Vf>>Vi, and Vf=nRT/pex, then w=-pex x nRT/pex=-nRT (im assuming n is number of moles of CO2?). 1 mole of CaCO3 makes 1 mole of CO2, so plugging in numbers, I get 8.9kJ, although I dont use the 1 atm pressure at all
6 0
3 years ago
For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution
Ann [662]

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}

<u>a = 8.06 m/s²</u>

3 0
2 years ago
When two objects with electrical charges interact, which affect the strength of that interaction?
noname [10]
The two objects with electrical charges interact, which affect the strength of that interaction <span>amount of charge. The answer is letter A. The rest of the choices do not answer the question above.</span> 
6 0
3 years ago
A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
A speeding car collides with an unlucky bug flying across the road. Which explains why the impact doesn’t equally damage the car
velikii [3]
The bug was a lot smaller than the car, that's for sure. The car is bigger and sturdier, while the bug is smaller and frail. The bug is so frail, that rather that putting a dent in the car, it splatters all over the car. The bug is very damaged (obviously), while the car just needs a good wash.
8 0
3 years ago
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