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1 year ago

1. Three students are using sports equipment to model the relative motion of

Physics

2 answers:

Alex Ar [27]1 year ago

3 0

The basketball would be sun then baseball earth and finally golf ball moon.

djyliett [7]1 year ago

3 0

Answer:

The golf ball would be revolving around the baseball while the baseball would be revolving around the basketball.

Explanation:

The basketball would be represented as the Sun, the baseball would be represented as the Earth and the golf ball would be represented as the moon. Since the Moon revolves around the Earth and the Earth revolving around the Sun makes a perfect explaination for this question.

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Would you be doing more work by going up the stairs twice as fast?

True [87]

Yes you would be doing more work

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2 years ago

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How many calories are needed to raise the temperature of 2g of water 3 C?

olga nikolaevna [1]

Answer:

The number of calories needed is 6c.

Explanation:

The amount of energy $Q$ needed to raise the temperature $\Delta T$ of water of mass $m$ is

$Q = mC\Delta T$

where $C = 1cal/g\cdot C$ is the specific heat capacity of water.

Putting in numbers into equation (1), we get:

$Q = (2g)(1)(3^oC )\\$

$\boxed{Q = 6\:cal }$

which is the number of calories needed.

8 0

2 years ago

5) help plz I need help I don't know if I am right so if I am not can you plz help me

seraphim [82]

the answer is waning Gibbous

The Waning Gibbous is an intermediary Moon phase. It starts right after the Full Moon, and it lasts until the Third Quarter.

7 0

2 years ago

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What is the current in a circuit with a 4.5 V battery and a 9 Ω resistor?

horsena [70]

Answer:

V=I×R

4.5 = I×9

 I. = 4.5/9

 I. = 0.5 A

current is 0.5 A

3 0

1 year ago

Read 2 more answers

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

2 years ago