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OLga [1]
3 years ago
8

A cart travels down a ramp at an average speed of 5.00 centimeters/second. What is the speed of the cart in miles/hour? (Remembe

r there are 100 centimeters per meter, 1000 meters/kilometer, and 1.6 kilometer per mile.)
Physics
2 answers:
viva [34]3 years ago
7 0
Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.

( \frac{ 5 cm}{1 sec} ) *( \frac{1m }{100cm})*( \frac{1km}{1000m} )*( \frac{1mi}{1.6km} ) = 0.00003125 mi/sec \\  \\ ( \frac{0.00003125 mi}{1sec} )*( \frac{60sec}{1min} )*( \frac{60min}{1hr} )=0.1125 mi/hr
Tju [1.3M]3 years ago
3 0
34 is the answer fo your question

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What affects sounds at their source? MENTION DOPPLER EFFECT AND VIBRATIONS
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A train travels 225 km in 2.5 hours. What is the trains average speed
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An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
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Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

b)

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

d)

Spring Constant:

As w=\sqrt{k/m}

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

⇒ v= 8.78 m/s

f)

Maximum force will be exerted on the block when it is at maximum distance.

F= k* A   ( A is amplitude of oscillation)

F= 179.2 * 0.27 N

F= 48.4 N

5 0
3 years ago
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