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3 years ago

15

A ball is thrown horizontally from the top of a tower at 30m high, and lands 15m from its base. What is the ball's initial speed

?

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Answer:

This question requires gravitational acceleration,t=（2h/g）^ 1/2,Level the initial speed is 15/t.

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An ant moves at 5 cm/s. It takes her 2 minutes to cross a road. How wide is the road?

Alecsey [184]

so as this ant moves

5 cm every second you multiply 5 by 120 (60 per minute as there are 60 seconds in a minute)

this is 600 cm

or

6 meters

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3 years ago

A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 88.4 Hz, determine the first three

yulyashka [42]

Answer with Explanation:

We are given that

Fundamental frequency,f=88.4 Hz

Speed of sound,v=343 m/s

We have to find the first three overtones.

Tube is closed

The overtone of closed pipe is equal to odd number of fundamental frequency.

Therefore, the overtone of tuba

$f'=nf$

Where n=3,5,7,..

Substitute n=3

$f'=3\times 88.4=265.2Hz$

For second overtone

$f'=5\times 88.4=442Hz$

For third overtone

$f'=7\times 88.4=618.8Hz$

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4 years ago

An airplane has an average speed of 740 mph

crimeas [40]

Answer:

2590

Explanation:

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3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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3 years ago

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Ulleksa [173]

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3 years ago