3 years ago

A runner makes one lap around a 200m track of 25s. What were the runners average speed and average velocity?

1 answer:

5 0

Here is the answer. In order for us to get the speed, we are just going to use the formula speed is equal to distance over time. Given that the distance s 200m and the time is 25s, this would give us the answer of 8m/sec. Hope this answers your question.

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Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?

kupik [55]

Half life is 5700 years,that is 50%.

6.25% means 1/16 of the life is left.

T^1/16 =T^1/2*4.

5700 × 4 =22800 years

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3 years ago

Scientists described light as a wave because the results of many experiments with light demonstrated that light behaves as a wav

Ksju [112]

The answer is Data collected from many experiments could not be explained.

Explanation:

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3 years ago

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G (where g=9.8 m/s2). If an object’s mass is m=10. kg, what is its weight?

Semenov [28]

$F_{w} =m*g \\ F_{w} =10*9.8=98N$

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3 years ago

A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such

Leviafan [203]

Answer:

ω = 0.571 rad/s

Explanation:

given data

radius = 30 m

solution

we take here g = 9.8 m/s²

and g is express as

g = r × ω² ....................1

put here value and we get

9.8 = 30 × ω²

solve it we get

ω = 0.571 rad/s

5 0

3 years ago

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is

HACTEHA [7]

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

$mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m$

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

5 0

3 years ago