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Nadya [2.5K]
3 years ago
7

The factors that affects the gravitational force between two objects are

Physics
1 answer:
viva [34]3 years ago
8 0

-- The product of their two masses (no matter what
the individual masses happen to be),

-- The distance between the centers of mass of the two masses.

Nothing else affects the force of gravity between two objects,
and nothing in between them can 'shield' or block it.


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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th
-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = -13.125 m/s^2

Explanation:

By momentum conservation along x direction we will have

mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2

so we have

v_1 + v_2 = 2v

v_1 + v_2 = 4.68

also by energy conservation

\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J

\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17

(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)

(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83

21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83

2v_2^2 - 9.36v_2 + 2.12 = 0

by solving above equation we will have

v_1 = 4.44 m/s

v_2 = 0.24 m/s

Average acceleration on first part of the chunk is given as

a_1 = \frac{4.44 - 2.34}{0.16}

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = \frac{0.24 - 2.34}{0.16}

a_2 = -13.125 m/s^2

5 0
3 years ago
3. a) If the distance between two objects increases by a factor of 5 (5),
irina [24]

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

3 0
2 years ago
The force of the added water produces a torque on the dam. In a simple model, if the torque due to the water were enough to caus
Fittoniya [83]

Answer:

The appropriate response is "\tau=\frac{1}{6} PgLh^3". A further explanation is described below.

Explanation:

The torque (\tau) produced by the force on the dam will be:

⇒  d \tau=XdF

On applying integration both sides, we get

⇒  \tau = \int_{0}^{a}x pgL(h-x)dx

⇒     = pgL\int_{0}^{h}(h-x)dx

⇒     =pgL[\frac{h^3}{2} -\frac{h^3}{3} ]

⇒     =\frac{1}{6} PgLh^3

8 0
2 years ago
A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
Crazy boy [7]
A) Work energy relation;
 Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
 F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
 F = 900/0.04
    = 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
 F = Ma
Thus; a = F/m
             = 22500/(5×10^-3)
             = 4,500,000 m/s²
But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
6 0
3 years ago
A group of humans traveling in space discover a habitable planet. They settle down there and start populating it. Their populati
storchak [24]

Answer:

10 years

Explanation:

As you can understand from the question it is given that the planet is already filled to half of its capacity. Also the population doubles in 10 years. To fill up the planet completely the population needs to double only once. To do that only 10 years are required.

As it is mentioned there are no other factors affecting the growth rate, in 10years the planet will be filled to its carrying capacity.

6 0
2 years ago
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