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olga nikolaevna [1]
3 years ago
7

Equal forces that do not cause a change in an object's motion

Physics
1 answer:
ss7ja [257]3 years ago
6 0

Answer:

Balanced forces.

Explanation:

The three main forces that stop moving objects are friction, gravity and wind resistance. Equal forces acting in opposite directions are called balanced forces. Balanced forces acting on an object will not change the object's motion. When you add equal forces in opposite direction, the net force is zero.

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Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h
nekit [7.7K]

Answer:

2.62seconds

Explanation:

Speed is defined as the ratio of the distance covered by a body with respect to time.

Speed v = Distance (s)/Time (t)

For a traveling sound wave, the distance between the source of a sound and the reflector is '2S'.

Speed v = 2 × distance (S)/Time (T)

V = 2S/t

2S = vt

Given speed of the wave = 342m/s

Distance covered = 450m

t = 2S/v

t = (2×450)/343

t = 900/343

t = 2.62seconds

It will take him 2.62seconds for him to hear his own voice echo off of the wall.

5 0
3 years ago
Read 2 more answers
Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec
sammy [17]

Answer:

1) The maximum jump height is reached at A. 0.337s

2) The maximum center of mass height off of the ground is B. 1.64m

3) The time of flight is C. 0.834s

4) The distance of jump is B. 7.49m

Explanation:

First of all we need to decompose velocity in its rectangular components, so

v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s

1) We use, v_{fy}=v_{iy}-gt, as we clear it for t and using the fact that v_{fy}=0 at max height, we obtain t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s

2) We can use the formula y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2} for t=0.337s, so

y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m

3) We can use the formula y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find total time of fligth, so 0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66, as it is a second-grade polynomial, we find that its positive root is t=0.834s

4) Finally, we use x=v_{x}t=8.05m/s(0.834s)=6.71m, as it has an additional displacement of 0.77m due the leg extension we obtain,

x=6.71m+0.77m=7.48m, aprox x=7.49m

5 0
4 years ago
20. <br> All financial goals are long-term goals.<br> True<br> False
Firdavs [7]

Answer:

Hello todays answer should be False i am sorry if i am incorrect-Lucy

Explanation:

8 0
3 years ago
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The force pressing two stacked objects together is equal to the weight of the object on top. The friction between a piece of woo
34kurt

Answer: B. Concrete

Explanation:

Let N = reacting force pressing the bodies in context together (units in Newtons),

The question stated that the force pressing the two mounted/stacked objects together is equal to the weight of the object on top.

We need to start by finding the weight of the piece of wood.

friction is given by

f = μN

The value of f is 22.5,

and from the chart reference the coefficient of friction between wood and stone, μ is 0.30.

22.5 = 75. 0.30

Putting the values into the equation: 22.5 = 0.30N.

Divide both sides by 0.30 to find the value of N:

N= 22.5/0.3 = 75

Now that the piece of wood will be placed on another surface, its weight of 75 Newton is the force pressing the two bodies together.

To determine the new surface, you should find the new coefficient of friction by using the new value of the force of friction given 46.5:

46.5 = µ(75).

Divide both sides by 75 to isolate μ.

The refer chart also indicates that the coefficient of friction equals 0.62 between wood and concrete, so the new surface corresponding to 0.62 is the concrete, which is (B).

3 0
4 years ago
What are 3 types of frictions and when does each apply/9029244/29edc63f?utm_source=registration
Anon25 [30]
- Static friction: when an object is not moving
- Kinetic friction: if an object is moving
- Rolling friction: when there is rolling (wheel,..)
5 0
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