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Keith_Richards [23]

3 years ago

10

What one unit of cane sugar is made of

1 answer:

Molodets [167]3 years ago

4 0

12 carbon, 22 hydrogen, 11 oxygen

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Which celestial body would have the strongest gravitational pull on a satellite orbiting 100 km above its surface?

8090 [49]

According to the Law of Universal Gravitation, the gravitational force is directly proportional to the mass, and inversely proportional to the distance. In this problem, let's assume the celestial bodies to be restricted to the planets and the Sun. Since the distance is specified, the other factor would be the mass. Among all the celestial bodies, the Sun is the most massive. So, the Sun would cause the strongest gravitational pull to the satellite.

3 0

4 years ago

Read 2 more answers

8) Calculate the kinetic energy of a truck that has a mass of 6100 kg and is moving at 55 m/s.

Inessa [10]

Answer:

<h2>9,226,250 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

v is the velocity

m is the mass

From the question we have

$k = \frac{1}{2} \times 6100 \times {55}^{2} \\ = 3050 \times 3025 \\ = 9226250$

We have the final answer as

<h3>9,226,250 J</h3>

Hope this helps you

7 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

Read 2 more answers

HELPP ME IN PHYSICS +15 POINTS!! <br> But a right answer not links or I’ll report. -_-

frozen [14]

The answer shud be 31 . 14m

8 0

3 years ago

Energy from the Sun that bounces off the surface of the Earth and returns to space is called...

Reika [66]

D is the answer trust me have faith it’s for the glory of humanity

5 0

3 years ago