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4vir4ik [10]
3 years ago
7

Two forces, one four times as large as the other, pull in the same direction on a 10kg mass and impart to it an acceleration of

2.5 m/s2. If the smaller force is removed, what is the acceleration of the mass?
Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

The acceleration of the mass is 2 meters per square second.

Explanation:

By Newton's second law, we know that force (F), measured in newtons, is the product of mass (m), measured in kilograms, and net acceleration (a), measured in meters per square second. That is:

F = m\cdot a (1)

The initial force applied in the mass is:

F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)

F = 25\,N

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to \frac{4}{5} of the initial force. The acceleration of the mass is:

\frac{25\,N}{20\,N} = \frac{2.5\,\frac{m}{s^{2}} }{a}

a = 2\,\frac{m}{s^{2}}

The acceleration of the mass is 2 meters per square second.

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expeople1 [14]
False, you pass a light through a mixture If the light bounces off the particles, you will see the light shine through and you have a colloid mixture
6 0
3 years ago
During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla
Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of u=-40\ m/s

ball returned with speed of v=40\ m/s

average Force imparted by racquet on the ball is given by

F_{racquet}=\frac{m(v-u)}{\Delta t}

where m=mass\ of\ ball

\Delta t=time of contact of ball with racquet

F_{racquet}=\frac{m(40-(-40))}{\Delta t}

F_{racquet}=\frac{80m}{\Delta t}-----1

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

F_{player}=\frac{m(0-40)}{\Delta t}

F_{player}=\frac{-40m}{\Delta t}-----2

From 1 and 2 we get

F_{racquet}=-2F_{player}

Hence the magnitude of Force by racquet is twice the Force by player

5 0
3 years ago
A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
UkoKoshka [18]

Answer:

Explanation:

1) Hypermetropia (better known as Farsighted- this is why nearby objects seem blurry for him)

2) In such instances, image are typically formed farther from the near point

3) Such defects are quite common so there are common procedures such as using convex lens which can restore the sight to normal.

7 0
3 years ago
Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt
Tamiku [17]

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of  A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am  1 m away from B , the path difference will be

8 - 1 = 7 m  which is  odd multiple of 1 or λ /2

So path difference becomes odd multiple of  λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

8 0
3 years ago
Read 2 more answers
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