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3 years ago

An object with a mass of 25 kg object is moving at 30 m/s. What is the

Physics

1 answer:

Pani-rosa [81]3 years ago

7 0

Answer: 750 kgm/s

Explanation:

Mass of object = 25 kg

Speed by which object moves =30 m/s. Linear momentum of the object = ?

Since momentum refers to the quantity of motion of the moving object,

Linear momentum = Mass x Speed

= 25kg x 30m/s

= 750 kgm/s

Thus, the linear momentum of the object is 750 kgm/s

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2 years ago

Light travels at 3.0 × 108 m/s in a vacuum and slows to 2.0 × 108 m/s in glass. What is the index of refraction of glass?

Slav-nsk [51]

We are asked to solve for the index of refraction and the formula is n = c/v where "n" represents the index of refraction, "c" represents the speed of light in the vacuum while "v" represents the speed of another medium.
In the problem, we have the given values below:
c = 3 x 10^8 m/s
v = 2 x 10^8 m/s
n =?

Solving for n, we have the solution below:
n = 3x10^8 / 2x10^8
n = 1.5

The answer is 1.5 for the index of refraction.

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3 years ago

Read 2 more answers

What is the frequency of radiation whose wavelength is 11.5 a0 ?

irakobra [83]

Answer:

The frequency of radiation is $2.61 \times 10^{17} s^{-1}$

Explanation:

Given:

Wavelength $\lambda = 11.5 \times 10^{-10}$ m

Speed of light $c = 3 \times 10^{8}$ $\frac{m}{s}$

For finding the frequency of radiation,

$c = f \lambda$

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }$

$f = 2.61 \times 10^{17}$ $s^{-1}$

Therefore, the frequency of radiation is $2.61 \times 10^{17} s^{-1}$

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3 years ago

Calculate the acceleration of the car for each set of conditions using the formula a = (v2 – v1) / (t2 – t1) where v2 and v1 are

Arisa [49]

Answer: 1. 0.19

2. 0.33

3. 0.47

4. 0.62

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3 years ago

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A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I

melamori03 [73]

(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
$v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s$

(b) The initial speed of the skater is
$v_i = 4 m/s$
while the final speed is
$v_f = 5.3 m/s$
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
$a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2$

(c) The initial speed of the skater is
$v_i = 13.0 m/s$
while the final speed is
$v_f=0$
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
$2aS=v_f^2 -v_i^2$
from which we find
$a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2$
where the negative sign means it is a deceleration.

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3 years ago