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Marysya12 [62]
3 years ago
8

When a sound wave encounters a barrier, what happens?

Physics
1 answer:
Tema [17]3 years ago
8 0
Sound waves will either be reflected (echo effect) or absorbed (dissipated) depending upon the material make-up of the barrier
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Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied
liraira [26]

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

EF_x = 0; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

= - ( 888 lbs - 632 lbs)

= - 256 lbs

5 0
3 years ago
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
Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble
Neporo4naja [7]

Answer:

2.4 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v

0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v

0.12 kg m/s = (0.05 kg) v

v = 2.4 m/s

4 0
3 years ago
Can someone please help​
coldgirl [10]

Answer:

Making a quick cut left to intercept a pass

Explanation:

It takes more energe to do than running

7 0
2 years ago
Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.
Sergeu [11.5K]
The work done to pull the sled up to the hill is given by
W=Fd
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is W=1900 J and the distance through which the force is applied is d=400 m, so we can calculate the average force by re-arranging the previous equation and by using these data:
F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N
4 0
3 years ago
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