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3 years ago

A stretched rubber band has _________energy

Physics

2 answers:

Mkey [24]3 years ago

6 0

Answer:

A stretched rubber band has potential energy

never [62]3 years ago

5 0

Answer:

elastic potential energy

Explanation:

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Which of the following is not conserved?

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B.) acceleration.
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3 years ago

The table represents a linear function. The rate of change between the points (–5, 10) and (–4, 5) is –5. What is the rate of ch

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4 years ago

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Temperature is____ to the average kinetic energy of a gas.

blsea [12.9K]

Answer: Option (b) is the correct answer.

Explanation:

As on increasing the temperature, the molecules gain more kinetic energy due to which they tend to collide and move rapidly from one place to another.

Thus, we can conclude that when temperature is increased, the kinetic energy of the molecules increases. This means that temperature is directly proportional to the average kinetic energy of a gas.

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4 years ago

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A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the s

yan [13]

Answer:

The force is $F_1 = 400.8 \ N$

Explanation:

From the question we are told that

The first diameter is $d_1 = 4.0 \ cm = 0.04 \ m$

The second diameter is $d_2 = 8.0 \ cm = 0.08 \ m$

Generally the first area is

$A_1 = \pi * \frac{d^2_1 }{4}$

=> $A_1 = 3.142 * \frac{0.04^2}{4}$

=> $A_1 = 0.00126 \ m^2$

The second area is

$A_2 = \pi * \frac{d^2_2 }{4}$

$A_2 = 3.142 * \frac{0.08^2}{4}$

$A_2 = 0.00503 \ m^2$

For a hydraulic press the pressure at both end must be equal .

Generally pressure is mathematically represented as

$P = \frac{F}{A}$

$\frac{F_1}{A_1 } = \frac{F_2}{A_2 }$

=> $F_1 = \frac{1600}{0.00503} * 0.00126$

=> $F_1 = 400.8 \ N$

8 0

3 years ago

A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0

3 years ago