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S_A_V [24]
3 years ago
5

In an automobile collision, a 44-kilogram passenger moving at 15 meters per second is brought to rest by an air bag during a 0.1

0-second time interval. What is the magnitude of the average force exerted on the passenger during this time?
Physics
1 answer:
const2013 [10]3 years ago
8 0

Answer:

6,600N

Explanation:

According to second law of motion, Force = mass × acceleration

If acceleration = change in velocity/time = 15/0.10

Acceleration = 150m/s²

Given mass = 44kg

Force = 44× 150

Force = 6,600N

Magnitude of the average force exerted on the passenger during this time is 6,600N

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NASA communicates with the Space Shuttle and International Space Station using Ku-band microwave radio. Suppose NASA transmits a
jeyben [28]

Answer:

λ₁ = 2.50 10⁻² m,   λ₂ = 1.66 10⁻² m  

Explanation:

Microwave communication is very efficient because it does not have atmospheric interference, for which it is widely used and has been regulated to avoid interference, the ku band is in the range between 12 and 18 GHz.

Let's calculate the wavelength for the two extreme frequencies of this band

wavelength and frequency are related

         c = λ f

          λ = c / f

f₁ = 12 GHz = 12 10⁹ Hz

          λ₁ = 3 10⁸ /12 10⁹

           λ₁ = 2.50 10⁻² m

f₂ = 18 GHz = 18 10⁹ Hz

          λ₂ = 3 10⁸ /18 10⁹

          λ₂ = 1.66 10⁻² m

Unfortunately in your exercise the specific frequency is not fired, for significant figures they must be the same number as the figures of the frequency, in general the frequency has 3 or 4 significant figures

8 0
3 years ago
Why is endurance training important to you?
Drupady [299]

Answer:

keeps your circulatory system healthy

Explanation:

7 0
3 years ago
Read 2 more answers
A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid
tatyana61 [14]

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

 we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = \frac{ r x ω^{2} }{g}

where ω (angular velocity) = \frac{2π}{time}

= \frac{2π }{5.5} = 1.14

Uk = \frac{ 5.7 x 1.14^{2} }{9.8} = 0.76

6 0
4 years ago
Which is longer, 10 cm or .01 m?
motikmotik

Answer:

They´re the same.

Explanation:

Someone deleted my answer. And my brainly is gone..

Have a good night ma´am/sir.

Be safe!

6 0
3 years ago
An object has a mass of 10grams and a volume of 20m3. what is the density of the object​
Pavlova-9 [17]

Explanation:

density = mass/ volume

in the question the mass is given in grams so will convert it into kg,

10 g = 0.01 kg

density= 0.01/ 20 = 1 / 2000 = 0.005 kg/m^3

7 0
3 years ago
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