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Novay_Z [31]
3 years ago
12

How is amplitude related to loudness

Physics
1 answer:
expeople1 [14]3 years ago
3 0
The loudness<span> of a sound is linked to the size of the vibration which produces it. A big vibration makes a louder sound. Scientists use the word '</span>amplitude<span>' for the size of waves. For waves on water, it is easy to measure the </span>amplitude<span>.</span>
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A stunt man jumps from the top of a building and lands 10 meters below his initial height. In case A, the stunt man lands on a s
krek1111 [17]

Answer:

Explanation:

The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.

Impulse=F_{avg}\times \Delta T

As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.  

8 0
3 years ago
How much net force is required to accelerate a 2000 kg car at 3.00 m/s^2
andrezito [222]

The net force required to accelerate a car is 6000 N.

Force is defined as the product of the mass and acceleration of the body. Force is used to changing the velocity that is to accelerate an object or a body of a particular mass. The unit of Force is Newton or kg m/s^2.

The formula used to calculate the net force is :

F = ma

where, F = Force

m = mass = 2000 kg

a = acceleration = 3.00 m/s^2

∴ F = 2000*3

F = 6000 N

Thus, to accelerate the car at 3.00 m/s^2 of mass 2000 kg net force required is 6000 N.

To learn more about force,

brainly.com/question/1046166

6 0
1 year ago
2.
anzhelika [568]

Answer:

44.72m/s

Explanation:

use th formula:vf²=vi²at

and then substitute the values

remember the units

8 0
2 years ago
In which of the following wave types is the particle motion perpendicular to wave motion?
galben [10]

Answer:

C...................

4 0
3 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
sergejj [24]

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

s_A = -6 + 9t

s_B = 5 - 8t

When A an B meets, they are at the same position and at the same time. So

s_A = s_B

-6 +9t = 5 - 8t

17t = 5 + 6 = 11

t = 11/17 = 0.647 s

s_A = -6 + 9*0.647 = -0.176 m

So where they meet is 0.176m from the flagpole, westward.

5 0
3 years ago
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