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NikAS [45]
3 years ago
7

The earth exerts a force of 1.00 N on an object in free fall. What is the object’s mass?

Physics
1 answer:
sp2606 [1]3 years ago
7 0

Answer: 0.1kg

Explanation:

Force=1N

Acceleration due to gravity(g)=10m/s^2

Mass=force/g

Mass=1/10

Mass=0.1kg

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A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at th
Shalnov [3]
Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.= 
<span>Wt. of skier. </span>

<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
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<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
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<span>Ek + Ep = Ekmax - Fk*d </span>
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4 0
4 years ago
Read 2 more answers
A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma
goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

F=ma

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

F=ma

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

x(t)=\frac{1}{2}a_g t^2

where a_g = 0.098 m/s^2 is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

x'(t)=15-\frac{1}{2}a_s t^2

where a_s=0.8 m/s^2 is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s

4 0
4 years ago
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