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3 years ago

6

At a baseball game, Bill observes from the bleachers Jack throwing a ball toward home at 15 km/s while Kevin runs toward home fr

om third base at 7 km/s. What is the velocity of the ball from Bill's and Jack's perspectives?

1 answer:

sattari [20]3 years ago

7 0

Answer:bill 5 m/s. Jack:10 m/s

Explanation:

Cuz I took it

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What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?

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The kinetic energy K given to the helium nucleus is equal to its potential energy, which is
$E=q \Delta V$
where q=2e is the charge of the helium nucleus, and $\Delta V$ is the potential difference applied to it.
Since we know the kinetic energy, we have
$E=K=85~keV=q \Delta V$
and from this we can find the potential difference:
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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

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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.

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Bettina spoke into a microphone during the school play to increase the sound of her voice so the audience could hear her speak.

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The loudness was increased by the amplifier which converted electrical energy into sound energy.

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To solve the problem we must know about the relationship between Speed, distance, and Time.

<h3>What is the relationship between Speed, distance, and Time?</h3>

We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

$\rm{Speed = \dfrac{Distance}{Time}$

The speed of the light is 30,060.12 km/sec.

Given to us

The distance between the earth and the sun is 15000000km
Light takes 499 seconds to reach earth from the sun.

We know that speed can be described as,

$\rm{Speed = \dfrac{Distance}{Time}$

Therefore,

<h3>What is the speed of the light?</h3>

$\text{Speed of light} = \dfrac{\text{Distance between the earth and the sun}}{\text{Time taken by the light to travel the distance}}$

Substitute the value,

$\text{Speed of light} = \dfrac{15,000,000\ km}{499\ seconds}$

$\text{Speed of light} = 30,060.12\ km/sec$

Hence, the speed of the light is 30,060.12 km/sec.

Learn more about Speed, distance, and Time:

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