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2 years ago

7

Describe You toss a ball to someone one meter away. Then you toss it to someone four meters away. How does your throw change?

1 answer:

krek1111 [17]2 years ago

3 0

We have that the Throw has changed in distance the ball has traveled and the Force applied in trowing the ball and Possibly the time of travel

From the Question we are told that

You toss a ball to someone one meter away

You toss it to someone four meters away

Generally

When you toss toss a ball to someone one meter away and You toss it to someone four meters away the are things that have changed

Force
Distance

Therefore

The Throw has changed in distance the ball has traveled and the Force applied in trowing the ball and Possibly the time of travel

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The musical tone a piano has a frequency of 346HZ and a wavelength of 1.3 M what is the speed of the sound

prohojiy [21]

Answer:

v= 449.8 m/s

Explanation:

Given data

Frequency= 346Hz

Wave length= 1.4m

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$v= f \lambda\\\\$

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2 years ago

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

bearhunter [10]

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

Total Energy before fall = Total Energy after fall

$E_{i}$ = $E_{f}$

Here, total energy is the sum of kinetic energy and potential energy

$K.E_{i}$ + $P.E_{i}$ = $K.E_{f}$ + $P.E_{f}$ (a)

When banana is at height h, it has

$K.E_{i}$ = 0 and $P.E_{i}$ = mgh

and when it reaches the river, it has

$K.E_{f}$ = 1/2m $v^{2}$ and $P.E_{f}$ = 0

Putting the values in equation (a)

0 + mgh = 1/2m $v^{2}$ + 0

mgh = 1/2m $v^{2}$

<em>cutting 'm' from both sides</em>

<em> </em>gh = 1/2 $v^{2}$

v = $\sqrt{2gh}$

Hence, the velocity of banana before hitting the water is

v = $\sqrt{2gh}$

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2 years ago

If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr

Agata [3.3K]

Answer:

Resistance in the flash tube, $R=3.97\times 10^{-3}\ \Omega$

Explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance, $C=503\ \mu F=503\times 10^{-6}\ F$

The time constant of RC circuit is given by :

$\tau=RC$

R is the resistance in the flash tube

$R=\dfrac{\tau}{C}$ ..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

$v=\dfrac{d}{\tau}$

$\tau=\dfrac{d}{v}$

$\tau=\dfrac{0.001}{500}$

$\tau=0.000002\ s$

Equation (1) becomes :

$R=\dfrac{0.000002}{503\times 10^{-6}}$

$R=3.97\times 10^{-3}\ \Omega$

So, the resistance in the flash tube is $3.97\times 10^{-3}\ \Omega$ . Hence, this is the required solution.

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2 years ago

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