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12345 [234]
3 years ago
7

g Can a rigid body experience any ACCELERATION when the resultant force acting on that rigid body is zero? Explain.Can a rigid b

ody experience any ACCELERATION when the resultant force acting on that rigid body is zero? Explain.
Physics
1 answer:
Advocard [28]3 years ago
5 0

Answer:

<em>No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.</em>

Explanation:

If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)

For a body under this type of equilibrium,

ΣF = 0   ...1

where ΣF is the resultant force (total effective force due to all the forces acting on the body)

For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as

ΣF = ma   ...2

where m is the mass of the body

a is the acceleration of the body

Substituting equation 2 into equation 1, we have

0 = ma

therefore,

a = 0

this means that<em> if the resultant force acting on a rigid body is zero, then there won't be any force available to produce acceleration on the body.</em>

<em></em>

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The lens in the eyepiece of a reflecting telescope breaks. How will this most affect the function of the telescope?
Serggg [28]

Answer:

B

Magnified images will not be created.

Explanation:

I did it and this was the correct answer

5 0
3 years ago
Which remains the same as the distance of an object from Earth changes?
Gemiola [76]

Answer:

MASS

Explanation:

weight has to do with the gravitational pull, as distance from earth decreases, so does the gravitational pull. meaning, the size of the force, the pull, and the weight would all change. mass stays the same (in this sense. if you gain weight on earth you will gain mass as well, but if you leave earth your weight will lessen and your mass will stay the same.)

hope this helped.

5 0
4 years ago
Read 2 more answers
Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the drawstring ba
Alex

Answer:

 v=39.05 m/s

Explanation:

Given that

x= 56 cm

F= 158 N

m= 58 g = 0.058 kg

Lets take spring constant = k

At the initial position,before releasing the arrow

F= k x

By putting the values

F= k x

158= 0.56 k

k=282.14 N/m

Now from energy conservation

Lets take final speed of the arrow after releasing

\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2

k x²=mv²

282.14 x 0.56² = 0.058 v²

v=39.05 m/s

3 0
3 years ago
A cosmic ray proton moving toward the Earth at 5.00 10 m/s 7 × experiences a magnetic force of 1.70 10 N −16 × . What is the str
EastWind [94]

Answer:

Magnetic field will be equal to B=3\times 10^{-5}T

Explanation:

We have given velocity of proton 5\times 10^7m/sec

Magnetic force experienced by proton F=1.7\times 10^{-16}N

Charge on proton q=1.6\times 10^{-19}C

Angle between field and velocity \Theta =45^{\circ}

Force in magnetic field is equal to F=qBVsin\Theta

So 1.7\times 10^{-16}=1.6\times 10^{-19}\times 5\times 10^7\times B\times sin45^{\circ}

B=3\times 10^{-5}T

So magnetic field will be equal to B=3\times 10^{-5}T

7 0
3 years ago
High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ba
pav-90 [236]

The question is incomplete. The complete question is :

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a  50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.

Solution :

We know that momentum = mass x velocity

The momentum of the golf club before impact = 0.200 x 60

                                                                             = 12 kg m/s

The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.

Now the momentum of the club after the impact is = 0.2 x 40

                                                                                    = 8 kg m/s

Therefore the momentum of the ball is = 12 - 8

                                                                = 4 kg m/s

We know momentum of the ball, p = mass x velocity

                                                     4 = 0.050 x velocity

∴ Velocity =  $\frac{4}{0.050}$

                 = 80 m/s

Hence the speed of the golf ball after the impact is 80 m/s.

6 0
3 years ago
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