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kow [346]
2 years ago
5

Which FBD would represent a car moving right with a motor force of 250 N, and force of friction of 750N, a weight of 8500N and a

normal force of 8500N
Physics
1 answer:
babymother [125]2 years ago
8 0

Answer:

Option C

Explanation:

Given that

Motor force is 250 N

Force of friction is 750 N

Weight is 8500 N

And, the normal force is 8500 N

Now based on the above information

Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force

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Answer:

E=6.8Kv/m

Explanation:

From the question we are told that

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Generally the equation for electric field at a distance is mathematically given as

E=\frac{v}{d}

E=\frac{533}{7.85*10^-^2}

E=6789.808917

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I think is 30m/s I am not sure
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A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose
dusya [7]

Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m  { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

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ahrayia [7]

Answer:

3430000 J

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From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).

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Velocity has two pieces of information.what are they
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