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nikklg [1K]
3 years ago
10

2. (Work-Energy Theorem) A watermelon of mass m is dropped from rest from the roof of a building of height h and feels no air re

sistance. a. What is the work done by gravity on the watermelon from the roof to the ground? b. What is the kinetic energy of the watermelon just as it hits the ground? c. What is the speed of the watermelon just as it hits the ground? d. Which answer(s) above will change if there is appreciable air resistance? m = 4.80 kg h = 25.0 m​
Physics
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Answer:

SORRY SISO REALLY DON'T KNOW ANS SORRY!........☹☹☹

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The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
E=U_{max}= \frac{1}{2}k(x_{max})^2

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
E=K_{max}= \frac{1}{2}m(v_{max})^2

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
U_{max}=K_{max}
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3 0
3 years ago
Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in t
zavuch27 [327]

This question is incomplete, the complete question is;

The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters

a) Find the volume V in terms of x and L

b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.  

Answer:

a) the volume V in terms of x and L is  ((√3/4)x²L) m³

b) required expression is (2/(3)^(1/u))√(At/L)

Explanation:

Given that;

form the question and image below;

triangular prism ends are equilateral triangle

side length = x meter

Dimension of the prism = L meter

Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter

Volume of the triangular prism = Area × height

= √3/4 (x)² × L

V = ((√3/4)x²L) m³

Therefore, the volume V in terms of x and L is  ((√3/4)x²L) m³

b)

Rate of water entering = A m³/hr

Depth of water tank = d meter

Time = t

Length of prism = L

now Rate of water entering is A m³/hr

dv/d = A                             [  V = ((√3/4)x²L) m³ ]

and

dv/dt = √3/4 [2x dx/dt ] L                   { L is constant }

so

A = √3/4 [2x dx/dt ] L  

∫A dt = √3/2 [ Lx dx ]                   { Integrate both sides}

At = √3/2 × Lx × x²/2

x² = uAt / √3L                              { we find square root of both sides}

x = √( uAt / √3L )

x = (2/(3)^(1/u))√(At/L)

Therefore; required expression is (2/(3)^(1/u))√(At/L)

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Which statement about electromagnetic waves is true?
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Answer:

B

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