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Arlecino [84]
4 years ago
8

A catapult’s lever holds a cannonball. The lever is attached to a tightly held rope. When the rope is released, the lever spring

s forward and launches the cannonball. When the rope is held tightly, which form of energy does it possess?
A. thermal
B. nuclear
(NOT) C. gravitational
D. elastic
Physics
2 answers:
PIT_PIT [208]4 years ago
8 0
The answer is elastic 
vlada-n [284]4 years ago
6 0
The answer is ELASTIC
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A rock is dropped in a pond, causing circular ripples. The radius increased with a rate 1 foot per second. When the radius is 4
Brrunno [24]

Answer:

true

Explanation:

Area = πr²

Differentiating in respect to r

we get,

dA/dr = 2πr

using chain rule,

\frac{dA}{dt} = \frac{dA}{dr} .\frac{dr}{dt}  = 2\pi \frac{dr}{dt}

since r = 4feet

so ,

2π(4) = 8π

5 0
3 years ago
A jet lands with a speed of 100 m/s and can accelerate uniformly at a rate of -5.0 m/s^2 as it comes to rest. What is the minimu
schepotkina [342]

Answer:

A ) 1000 m.

Explanation:

Here initial velocity u = 100 m /s

Final velocity v = 0

Acceleration a = -5 ms⁻²

Distance travelled = S

v² = u² + 2aS

0 = (100)² -2 x 5 S

S = 10000/ 10

=1000 m.

3 0
3 years ago
A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

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