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Ilya [14]
3 years ago
6

A girl, moving at 8.8 m/s on in-line skates, is overtaking a boy moving at 4.3 m/s as they both skate on a straight path. The bo

y tosses a ball backward toward the girl, giving it speed 11.4 m/s relative to him. What is the speed of the ball relative to the girl, who catches it?
Physics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:16.5 m/s

Explanation:

Given

Velocity of boy(v_b)=4.3 m/s

velocity of girl(v_g)=8.8 m/s

velocity of ball relative to boy(v_{Bb})=11.4 m/s backward

Now We have to find velocity of ball relative to girl

Representing it in vector form

v_b=4.3\hat{i}

v_g=8.8\hat{i}

v_{Bb}=-11.4\hat{i}

and

v_{Bb}=v_B-v_b

v_B=v_{Bb}+v_b

v_B=-11.4\hat{i}+4.3\hat{i}=-7.1\hat{i}

therefore velocity of ball relative to girl

v_{Bg}=v_B-v_g=-7.1\hat{i}-8.8\hat{i}=-16.5\hat{i}

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Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

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the changes in waves

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3 years ago
A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times
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Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

            L=10log_{10}\left ( \frac{I}{I_0}\right )

A train whistle has a sound intensity level of 70 dB

We have

           70=10log_{10}\left ( \frac{I_1}{I_0}\right )

A library has a sound intensity level of about 40 dB

We also have

           40=10log_{10}\left ( \frac{I_2}{I_0}\right )

Dividing both equations

           \frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000

The sound intensity of train is 1000 times greater than that of the library.

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2 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
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Explanation:

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Focal length of the converging lens, f = 0.14 m

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\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

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Examples of energy balance: physical education
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