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3 years ago

What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?

Physics

1 answer:

Klio2033 [76]3 years ago

4 0

Force =mass x acceleration =1100 x0.5 =550

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Assume an 8-kg bowling ball moving at 2m/s bounces off a spring at the same speed that it had before bouncing.What is the moment

aliya0001 [1]

Answer:-16kgm/s

Explanation:

-mv=-2×8=-16kgm/s

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3 years ago

The result of a wave generator traveling faster than the speed of a wave is:

Vesnalui [34]

The result of a wave generator traveling faster than the speed of a wave is called as a boom. If the wave is a sound wave, it is called a sonic boom. However, if the wave is light, it is called as a luminal boom. Luminal bloom happens in some industries and is commonly called as the Cherenkov radiation.

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3 years ago

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In " m a x f o r t e d u c a t i o n " . what is probability of vowels.

LiRa [457]

Answer:

7/16 is the probability of given querty

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3 years ago

A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards

Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

$Y = 1.6 X - 0.032x^2$

we know that maximum height occurs is given as

$x = -\frac{b}{2a}$

$y =- \frac{1.6}{2(-0.032)} = 25$

and maximum height is

$y = 1.6\times 25 - 0.032\times 25^2$

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

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3 years ago

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A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl

raketka [301]

The wavelength of the note is $\lambda = 39.1 cm = 0.391 m$ . Since the speed of the wave is the speed of sound, $c=344 m/s$ , the frequency of the note is
$f= \frac{c}{\lambda}=879.8 Hz$

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where $\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m$ is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
$L= \frac{1}{2f} \sqrt{ \frac{T}{\mu} } =0.31 m$

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3 years ago