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2 years ago

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Wat would happen if a feather and a ball were released from the same height at the same time? Gravity Experiments for Kids – Sci

ence Experiments for Kids

1 answer:

solniwko [45]2 years ago

6 0

Answer:

They would land at the same time

Explanation:

They would land at the same exact time.

As weird, impossible and unbelievable as it appears. When in a vacuum, every weight, body and material when released from the same height would land on the ground at the same time. This also means that like in the question, a feather and a ball would land at the same time. And just for illustrations as well, a feather and a car would land at the same time as well.

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Physics question, any help is appreciated :)

Gekata [30.6K]

Runner 2 sees Runner 1 passing him with a velocity of 17 m/s west.

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2 years ago

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How do forest fires affect the Earth's atmosphere?

netineya [11]

It is B because the other ones are good.

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2 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

7 0

2 years ago

How magnitude, distance and velocity affects motion?

levacccp [35]

Answer:

Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

Explanation:

Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

6 0

2 years ago

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$ , $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

$B = 5.33\times10^{-5} T$

Therefore, the magnitude of magnetic field at given point = $5.33\times10^{-5} T$

3 0

3 years ago