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Kitty [74]
2 years ago
12

o illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using t

he work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone.
Physics
1 answer:
Phantasy [73]2 years ago
5 0

Answer: force of gravity on the body due to height difference above the earth's surface

Explanation: as you increase the height of a body above ground, you do work against gravity in moving it from a point on the earth's surface to that point. So a body falling has a stored up gravito-potential energy which acts on it downward due to its mass, accelerating it downwards

Answer b): kinetic energy of the body

Explanation: the downward force produces an acceleration of magnitude 9.81m/s2 downwards which means an increasing velocity. This increasing velocity means the kinetic energy of the body is increasing (kinetic energy is proportional to velocity of the body squared)

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A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

8 0
2 years ago
A cylinder of radius R and height H is floating upright in
emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

F = mg

so we have

\Delta P \pi R^2 = mg

so we have

\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g

so we have

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

6 0
2 years ago
When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
Degger [83]

Answer:

(C) 4 beats per second.

Explanation:

As we know that the no of beats can be calculated as.

No. of beats is equal to difference in the tuning forks frequencies.

So,

n= \nu _{1}- \nu _{2}.

Substitute the values of frequencies of 2 tuning forks in the above equation.

n=(260 Hz-256 Hz)\\n=4

Therefore the number of beats per second will be hear by the observer is 4 beats per second.

3 0
2 years ago
Wood block 1 in (Figure 1), which has a mass of 1.0 kg, is at rest on a wood ramp. The angle of the ramp is 20º above horizontal
Sergeeva-Olga [200]

Answer:

Explanation:

For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.

X axis

      -Aₓ - f_e +T = 0        (1)

Y axis

      N₁ - W_y = 0              ( 2)

let's use trigonometry for the weight components

      sin θ = Wₓ / W

      cos θ = W_y / W

      Wₓ = W sin θ

      W_y = W cos θ

We write the diagram for the second body.

Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards

      W₂ -T = 0                             (3)

we add the equations is 1 and 3

       - W₁ sin θ - μ N₁ + W₂ = 0

from equation 2

       N₁ = W₁ cos θ

       

we substitute

        -W₁ sin θ - μ (W₁ cos θ) + W₂ = 0

W₂ = m₁ g (without ea - very expensive)

This is the smallest value that supports the equilibrium system

7 0
2 years ago
Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
ioda

Answer:

6.1875\times 10^{-8}

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies I=\frac {k}{r^{2}} where r is the distance

Since intensity is given then when at 3 m

1.1\times 10^{-7}= \frac {k}{3^{2}}

k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}

Since we have the constant then at 4m

Intensity, I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}

8 0
1 year ago
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