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worty [1.4K]
3 years ago
12

For a standing wave to form in a medium, two waves must

Physics
1 answer:
Slav-nsk [51]3 years ago
4 0
For a standing wave to form, two waves must be traveling in opposite directions and cause destructive interference.
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Which of the following is supportive evidence of the big bang theory? pulsars the different shapes of galaxies cosmic microwave
Lunna [17]

Cosmic microwave background radiation

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Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction
Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity  is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension  force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0  

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁     ,    T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0
3 years ago
1. If you push a 40.0kg mass object with a force of 145.ON, what will be the object's
horrorfan [7]

Newton said . . . F = m a

Divide each side by  m . . .  a = F / m

Acceleration = (force) / (mass)

Acceleration = (145.O N) / (40.0 kg)

<em>Acceleration = 3.625 m/s²</em>

4 0
3 years ago
A 23 kg child is riding a 6.7 kg bike with a velocity of 4.4 m/s to the northwest.
mash [69]
How am I supposed to know the momentum of anything ? What does it matter
5 0
3 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
marta [7]

1) Acceleration of the car in front: -7.89 m/s^2

The only data we need for this part of the problem is:

u = 27.7 m/s --> initial velocity of the car

\mu=0.804 --> coefficient of friction between the car wheels and the road

From the coefficient of friction, we can find the deceleration of the car. In fact, the force of friction is given by

F=-\mu mg

where m is the car's mass and g=9.81 m/s^2 is the acceleration due to gravity. We can find the car's acceleration by using Newton's second law:

a=\frac{F}{m}=\frac{-\mu mg}{m}=\mu g=(0.804)(9.81 m/s^2)=-7.89 m/s^2

And the negative sign means it is a deceleration.


2) Braking distance for the car in front: 48.6 m

This can be found by using the following SUVAT equation:

v^2 - u^2 = 2aS

where

v=0 is the final velocity of the car

u=27.7 m/s is the initial velocity of the car

a=-7.89 m/s^2 is the acceleration of the car

S is the braking distance

By re-arranging the formula, we find S:

S=\frac{v^2-u^2}{2a}=\frac{0-(27.7 m/s)^2}{2(-7.89 m/s^2)}=48.6 m


3) Minimum safe distance at which you can follow the car: 15.0 m

In this case, we must calculate the thinking distance, which is the distance you travel before hitting the brakes. During this time, the speed of your car is constant, so the thinking distance is given by

d_t = ut=(27.7 m/s)(0.543 s)=15.0 m

After hitting the brakes, your car decelerates at the same rate of the car in front of you, so the braking distance is the same of the other car:

d_b=48.6 m

So the total distance your car covers is

S'=d_t+d_b=15.0 m +48.6 m=63.6 m

At the same time, the car in front of you just covered a distance of 48.6 m. So, in order to avoid the collision, you should travel at a distance equal to

d=S'-S=63.6 m-48.6 m=15.0 m


6 0
4 years ago
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