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muminat
3 years ago
8

Some believe that the positions of the planets at the time

Physics
1 answer:
laila [671]3 years ago
7 0

Answer:

1.4007\times 10^{-8}\ N

1.50075\times 10^{-6}\ N

0.000000667\ N

Explanation:

m_1 = Mass of baby = 3 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Distance between objects

Gravitational force of attraction is given by

F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 70\times 3}{1^2}\\\Rightarrow F=1.4007\times 10^{-8}\ N

The force between baby and obstetrician is 1.4007\times 10^{-8}\ N

F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(6\times 10^{11})^2}\\\Rightarrow F=1.50075\times 10^{-6}\ N

The force between the baby and Jupiter is 1.50075\times 10^{-6}\ N

F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(9\times 10^{11})^2}\\\Rightarrow F=0.000000667\ N

The force between the baby and Jupiter is 0.000000667\ N

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A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it
anzhelika [568]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = \frac{sum of coefficient of static friction}{number of trials}

                                              = \frac{0.779}{6}

                                              = 0.12983

The average coefficient of static friction is 0.130

8 0
3 years ago
Can you explain why number 3 is correct?
lukranit [14]
No waves because Q19 waves would going at the surface at regions
7 0
3 years ago
A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-forme
Sindrei [870]
I need to know this answer
5 0
4 years ago
Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper
vladimir1956 [14]

Answer:

Total distance does the grasshopper travel before the cars hit is 150 m

Explanation:

Each car moves x=100 m before they collide. Both the cars moving in constant velocity. time taken t by each car is

t=\frac{x}{v}

where x  is the distance traveled with velocity v

t=\frac{100}{10}\\t=10 sec

The insect is moving through this time period with a constant velocity of 15 m/s

The distance traveled by grasshopper  is

distance=V_{gh} \times t\\distance=15 \times 10\\distance=150 m

7 0
3 years ago
If your friend said that kinetic energy was changing to potential energy at point C, how would you respond? A) your friend is co
Travka [436]

funny of u to assume I have friends

If I remember anything from that part of my education (not great at physics) I'd say the answer is A, though I admit i'm not 100% sure

I dunno how to explain, once it hits that's the energy that was converting I guess.

3 0
3 years ago
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