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stellarik [79]
3 years ago
15

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. The force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00 x 103 m/s, given the collision lasts 6.00 x 10-8 s is Fill input: x 106 N.
Physics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

The correct answer is "6666.67 N".

Explanation:

The given values are:

Mass,

m = 0.100

Relative speed,

v = 4.00 x 10³

time,

t = 6.00 x 10⁻⁸

As we know,

⇒  F=m(\frac{\Delta v}{\Delta t} )

On substituting the given values, we get

⇒      =0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )

⇒      =6666.67 \ N

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OlgaM077 [116]

Answer:

The magnification is a function of the lenses in the objective and the eyepiece, so the magnification of the two must be multiplied to obtain the total magnification possible. So, for example, if the objective lens was 4X and the eye piece lens was 10X, the total magnification would be 40. (4 x 10 = 40)

Explanation:

5 0
2 years ago
A particle with a charge of 4.4 * 10^-5 C is released in an electric field whose magnitude is 750 N/C. What is the force that th
padilas [110]

Answer:

F = 3.3×10^ -2 N

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2 years ago
How are mechanical waves and electromagnetic waves alike?
miss Akunina [59]

I got this answer from the internet

6 0
3 years ago
Read 2 more answers
A 6 cm diameter drill bit is used to drill a cylindrical hole through the middle of a sphere of radius 5 cm. What is the volume
ira [324]

Answer:

2.4086 * 10^(-4)  m^3

Explanation:

Volume of sphere = 4/3 * pi * r1^3

Volume of cylinder = pi*r2^2*h

r1 = 0.05 m

r2 = 0.03 m

Hence,

Resulting Volume = Volume of sphere - Volume of cylinder

= \frac{4}{3}*pi*(0.05)^3 - pi*(0.03)^2*(0.1)\\= 5.236 * 10^(-4)  - 2.8274*10^(-4) \\= 2.4086 * 10^(-4) m^3

3 0
3 years ago
If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?
Law Incorporation [45]

a) 4.8 m/s² his average acceleration during the first 2.5 s

b) 15 m distance he cover during the first 2.5 seconds.

c) 31 m/s his speed as he finished the race.

<h3>(a) How to calculate the Average Acceleration ?</h3>

Average acceleration is calculated by using Newtons first law of motion

v = u + at

where

u is initial velocity

v is final velocity

a is constant acceleration

t is time

Given u = 0m/s , v = 12 m/s and t = 2.5 sec

Therefore average acceleration is given by

a=\frac{v}{t}

a=\frac{12}{2.5}

a = 4.8 m/sec²

b) Here we will use newtons second law of motion

s=ut+\frac{1}{2}at^{2}

On substituting value we get

s = 15m

c) Here we will use newtons third law of motion

v² = u² + 2as

Here u = 0 , a = 4.8 m/s² and s = 100 m

Therefore

v² = 960

v = 31 m/s

Disclaimer: the question was given incomplete in the portal. Here is the complete question.

Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.

a. What was his average acceleration during the first 2.5 s?

b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?

c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

Learn more about Newtons law of motion here:

brainly.com/question/12525794

#SPJ4

5 0
2 years ago
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