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Igoryamba
3 years ago
13

What forces were exerted during the air resistance experiment?

Physics
1 answer:
zepelin [54]3 years ago
6 0

The forces that were exerted during the air resistance was the gravity.

<u>Explanation:</u>

Air resistance is the frictional power air applies against a moving article. As an article moves, air opposition backs it off. The quicker the article's movement, the more prominent the air obstruction applied against it.

The force acting upon the objects dropped in the air was air resistance and gravity. The force acting upon the objects dropped in the vacuum was gravity only.

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An 8 kilogram bowling ball is rolling in a straight line toward you. if its momentum is 16 kg* m/s, how fast is it traveling ?
morpeh [17]
From the information given, The mass of the bowling ball is 8 Kilograms and the momentum with which it is moving is 16 kg m/s. We use the formula p = m × v Where p is the momentum, m is the mass and v is the velocity. We need velocity so we rewrite the equation thus: P = mv, therefore p/m = v or v = p/m In our case p = 16 and m = 8 v = p/m v = 16/8 v = 2 Therefore the bowling ball is travelling at 2m/s
6 0
3 years ago
In what setting do all frequencies of electromagnetic waves travel at the same speed?
Irina-Kira [14]
All electromagnetic waves travel in vacuum (space) at the speed of light (3 * 10^8 m/s). Radio waves is just a member of the electromagnetic spectrum. All electromagnetic waves follow the wave equation: speed = frequency * wavelength. With all electromagnetic waves, the speed in space is the same.
6 0
3 years ago
Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.
Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

Q=-P_{tot}=-(P_{2}-P_{1})

The potential energy for any point can be calculated as:

P=m*g*h

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

Q=P_{1}=m*g*h

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW

5 0
3 years ago
An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the
Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0
3 years ago
A satellite orbits the earth a distance of 2.50 × 10^7 m above the planet’s surface and takes 6.43 hours for each revolution abo
ASHA 777 [7]

Answer:

2.30 m /sec2

Explanation:

See the attached file;

7 0
3 years ago
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