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Helga [31]
3 years ago
13

What happens to the velocity of an object when balanced forces act on it

Physics
1 answer:
mixas84 [53]3 years ago
6 0
Nothing happens to velocity at all. Speed and direction remain constant.
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A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th
Juliette [100K]

Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

= 0.0180701s

5 0
3 years ago
You are writing a report about space travel. Where would be the best place to find information? (2 points)
Harrizon [31]

Answer: Contact the National Aeronautics and Space Association

Explanation:

The National Aeronautics and Space Association is an independent organization in US which aims to enable a safe, secure, efficient space journey and launch of new satellites, spacecraft beyond the earth's orbit. It can give better information for the space travel already conducted by the astronauts. Thus will help in writing the report as it is the best place to find the information.  

5 0
3 years ago
Read 2 more answers
Two polarizers are oriented at 68 ∘ to one another. unpolarized light falls on them. part a what fraction of the light intensity
Radda [10]
Your answer is.07 hope this helped 
4 0
3 years ago
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A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in
Arada [10]

Answer:

5295.3 N

Explanation:

According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force

mv + P = mV

P = mV - mv = m(V - v)

Since the velocity of the ball before and after is in opposite direction, one of them is negative

P = 0.14(44.8 - (-19.5)) = 9 kg m/s

Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is

F = P/t = 9/0.0017 = 5295.3 N

4 0
3 years ago
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