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3 years ago

A diver leaps from the edge of a diving platform

Physics

1 answer:

Pepsi [2]3 years ago

7 0

Please reply to this with your question I would like to help.

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Question 8

Viefleur [7K]

Answer:

96 Joules

Explanation:

The formula for work is Fnet times displacement (F x d = w) which, in this case, 48N is the Fnet and 2m as the displacement. Then all we need to do is multiply these two and we get 96 Joules.

3 0

3 years ago

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

puteri [66]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

3 years ago

Which of the following is true about all electromagnetic waves in a vacuum

Helen [10]

Electro waves in a vacuum air is deals with this and electricity when the air and the electricity it makes electro magnets.

4 0

3 years ago

Read 2 more answers

Diagram shows the velocity time graph for a particle moving in 16 seconds.

Leya [2.2K]

Answer:

The detailed calculations are shown below;

Explanation:

a)The maximum acceleration of the particle:

It is seen that the maximum change in velocity is at the time between 8s to 10s.

Maximum acceleration: $\frac{V}{t}$

= $\frac{20}{2}$

= 10 m/ $s^{2}$

b) The deceleration of the particle

The velocity of particle is decreased after 10s so,

deceleration = - $\frac{40}{6}$

= - 6.67 m/ $s^{2}$

c)The total distance traveled by the particle = Area under the curve

= $\frac{1}{2}$ * 4*20 + 4*20 + $\frac{1}{2}$ * 2*20+ 2*20+ $\frac{1}{2}$ * 40*16

= 290 m

d)The average velocity of the particle = $\frac{Area under the curve}{Total time}$

= $\frac{290}{16}$

= 18.12 m/s

3 0

3 years ago

If a space probe sends back info on an object traveling through space at close to the speed of light astronomers will be working

OleMash [197]

The speed of light in a vacuum is 300,000,000 m/s
That is three hundred million meters per second.
Therefore, the answer is
C) millions of meters per second

8 0

3 years ago