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Serjik [45]
3 years ago
6

If the distance between two objects is decreased to 1 10 of the original distance, how will it change the force of attraction be

tween them? A. The new force will be 100 times more than the original. B. The new force will be 20 times more than the original. C. The new force will be 1 20 of the original. D. The new force will be 1 100 of the original. E. The new force will be 1 10 of the original.
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

(A) It will 100 times larger than the original force.

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A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp
alex41 [277]

Answer:

5 m/s2, left

Explanation:

We can solve the problem by applying Newton's second law of motion, which  states that:

\sum F=ma

where:

\sum F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have:

\sum F=30 N - 20 N = 10 N (to the left) is the net force on the object

m = 2.0 kg is the mass

So, the acceleration is:

a=\frac{\sum F}{m}=\frac{10}{2.0}=5.0 m/s^2

in the same direction as the force (left).

5 0
3 years ago
Decreasing the trains velocity will ______ the kinetic energy.<br><br> Decrease or Increase
alex41 [277]
Decreasing the trains velocity will DECREASE the kinetic energy
8 0
3 years ago
Please help me fast please help​
inysia [295]

Answer:

the net energy Gained per hour equals 30Kcal/h

4 0
3 years ago
A string is stretched to a length of 308 cm and both ends are fixed. If the density of the string is 0.023 g/cm, and its tension
soldi70 [24.7K]

Answer:

Frquency=3,994Hz

Explanation:

Tension =967N

Density of string (μ)=0.023g/cm

Length of the stretched spring=308cm

Fundamental frequency for nth harmonic :

Fn=n/2L(√T/μ)

Substituting the given values to find the frequency :

f1=1/2(308cm) *(0.01m/1cm)[(√967N)/(0.023g/cm)(0.1kg)/(0.1kg/m)/(1g/cm)]

=6.16m[(√967N)/0.0023kg/m)]

=3,994.20Hz

Approximately,

The frequency will be =3,994Hz

7 0
3 years ago
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