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Katena32 [7]
4 years ago
8

Two solid balls (one larger, the other small) and a cylinder roll down a hill. Which has the greatest speed at the bottom and wh

ich the least?
(a) the larger ball has the greatest, the small ball has the least.
(b) the small ball has the greatest, the larger ball has the least.
(c) the cylinder has the greatest, the small ball has the least.
(d) both balls have the same greater speed, the cylinder has the least.
Physics
1 answer:
makvit [3.9K]4 years ago
6 0
The answer is c hope this clarifies
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Some words that have changed meaning due to technological advances are dial, type, tweet, drone, and spam.

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Which body system is responsible for bringing oxygen into the body?
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The respiratory system
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3 years ago
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You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain too
steposvetlana [31]

Answer:

w'=5.679\ N on the planet

w=22.43\ N on earth

Explanation:

Given:

  • initial velocity of the tool before pushing, u=0\ m.s^{-1}
  • force applied on the tool, F=12\ N
  • displacement of the tool, s=16.4\ m
  • time taken for the displacement, t=2.5\ s
  • height of releasing  the tool, h=10.3\ m
  • time taken by the tool to fall on the ground, t_v=2.88\ s

<u>Now using the equation of motion:</u>

s=u.t+\frac{1}{2}a.t^2

where:

a = acceleration of the object

16.4=0+0.5\times a\times 2.5^2

a=5.248\ m.s^{-2}

Now the mass of the tool:

m=\frac{F}{a}

m=\frac{12}{5.248}

m=2.2866\ kg

<u>Using the equation of motion when the tool is dropped:</u>

h=u.t_v+\frac{1}{2} \times g.t_v^2

here:

g = acceleration due to gravity on the planet

10.3=0+0.5\times a\times 2.88^2

g=2.4836\ m.s^{-2}

Weight of the tool in the planet:

w'=m.g

w'=2.2866\times 2.4836

w'=5.679\ N

Weight of the tool on the earth:

w=m.g'

w=2.2866\times 9.81

w=22.43\ N

8 0
4 years ago
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Two waves have the same speed. the first has twice the frequency of the second. compare the wavelengths of the two waves. 1. the
guajiro [1.7K]
The relationship between wavelength, speed and frequency of a wave is given by
\lambda= \frac{v}{f}
where
\lambda is the wavelength
v the speed
f the frequency

For the first wave, we can write
\lambda_1 =  \frac{v}{f_1}
while for the second wave
\lambda_2 =  \frac{v}{f_2}
where v is the same for two waves, since they have same speed. The first wave has twice the frequency of the second, so
f_1 = 2 f_2
So we can rewrite the wavelength of the first wave as
\lambda_1 =  \frac{v}{2 f_2}= \frac{1}{2} \frac{v}{f_2}= \frac{1}{2}  \lambda_2

which means that the correct answer is
<span>3. the first has half the wavelength of the second</span>
5 0
3 years ago
In case 1, a block hanging on a spring oscillates with amplitude dd. in case 2, an identical block hanging on an identical sprin
Bogdan [553]

Answer:

period in case 2 is \sqrt{2} times the period in case 1

Explanation:

The period of oscillation of a spring is given by:

T=2 \pi \sqrt{\frac{m}{k}}

where

m is the mass hanging on the spring

k is the spring constant

Therefore, in order to compare the period of the two springs, we need to find their m/k ratio.

We know that when a mass hang on a spring, the weight of the mass corresponds to the elastic force that stretches the spring by a certain amplitude A:

mg = kA

So we find

\frac{m}{k}=\frac{A}{g}

The problem tells us that the amplitude of case 1 is d, while the amplitude in case 2 is 2d. So we can write:

- for case 1:

\frac{m}{k}=\frac{d}{g}

T_1=2\pi \sqrt{\frac{d}{g}}

- for case 2:

\frac{m}{k}=\frac{2d}{g}

T_2=2\pi \sqrt{\frac{2d}{g}}

And by comparing the two periods, we find:

\frac{T_2}{T_1}= \sqrt{2}

So, the period of oscillation in case 2 is \sqrt{2} times the period of oscillation in case 1.

4 0
3 years ago
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