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3 years ago

What was the average speed for the second leg of the trip? help me im stuck

Physics

1 answer:

SVEN [57.7K]3 years ago

3 0

Answer:

The average speed on the second leg of the trip is 2 m/s

Explanation:

Notice that the second leg of the trip is from 8 m in the horizontal axis heading to the left to position 4 m. So the velocity is going to be negative and given by the quotient of the change in position (-4 m) divided by the time it took: 2 seconds.

This gives: -4/ 2 m/s = -2 m/s

But the speed is going to be the absolute value of the velocity, that is:

2 m/s

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Answer:

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3 years ago

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To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.

Our values are given as

$f = 4.11 *10^{12} Hz$

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The angular velocity of a body can be described as a function of frequency as

$\omega = 2\pi f$

$\omega = 2\pi 4.11 *10^{12}$

$\omega=2.582*10^{13} rad/s$

PART A) The expression for the maximum angular velocity is given by the amplitude so that

$V = A\omega$

$V =( 1.23 * 10^{-11})(2.582*10^{13})$

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PART B) The maximum acceleration on your part would be given by the expression

$a = A \omega^2$

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3 years ago

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2 years ago

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A 15 g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 53

blagie [28]

Answer:

0.0613°C

Explanation:

the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s

the bullet moves with different kinetic energies before and after the penetration, therefore

Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)

= $\frac{1}{2}$ × 15×10⁻³ × (865² - 534²)

= 3.47 × 10⁻³J

this loss in energy is transferred to the water, therefore

change in temperature = $\frac{Q}{m C}$

where c = heat capacity of water = 4.19 x 10^3

m = mass of water = 13.5 kg

= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }

=0.0613°C

5 0

3 years ago