All categories

1 year ago

How does position depend on time on a free falling motion, for short distance, near the surface of the earth?

Physics

1 answer:

Korvikt [17]1 year ago

7 0

Answer:

if an object is in free fall for a longer time its position will increase at an exponential rate because the object is accelerating

You might be interested in

The place kicker on a football team kicks a ball from ground level with an initial speed of 4.00 m/s at an angle of 29.0° above

mafiozo [28]

Answer:

0.4 s

Explanation:

The time that the ball is in the air can be found with the next formula:

$time = t = \frac{2v_{o}sen\beta}{g}$

where $v_{o}$ is the initial velocity of the ball, in this case:

$v_{o}=4.00 m/s$

β is the angle: β = 29.0°

And g is the acceleration of gravity: $g=9.81m/s^2$

Replacing all the values to find t, we have:

$time = t = \frac{2(4m/s)sen(29)}{9.81m/s^2} = 0.4s$

Thus, the ball is 0.4s in the air.

3 0

3 years ago

The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance

belka [17]

The gravitational force between the spheres is

$F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N$

where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².

The weight of the lighter sphere is

$F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N$

where <em>g</em> = 9.80 m/s².

The ratio between the two forces is then

$\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}$

7 0

3 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What is the new vol

Alborosie

Initial volume of the gas (V1) = 10 inches^3
Initial pressure (P1) = 5 psi
Final pressure after compression of the gas (P1) = 10 psi
Let us assume the final volume of the gas (V2) = x
According to Boyle's Gas law, the pressure and volume of a gas remains constant under ideal condition. Then
P1V1= P2V2
5 * 10 = 10 * x
50 = 10x
x = 50/10
= 5 cubic inches
So the volume of the gas after it was compressed was 5 cubic inches. I hope the procedure is clear enough for you to understand.

8 0

3 years ago

Unmoving objects are being acted on by balance forces. TRUE or FALSE?

Reil [10]

True. There are forced acting on it, but as they're balanced it is unmoving

3 0

3 years ago

Read 2 more answers

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

bezimeni [28]

Answer:

a) The resulting angular speed of platform is 1.38 rev/sec

b) The change in kinetic energy of the system is 53 J.

Explanation:

This question is incomplete. The complete question will be:

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 7.0 k g .m 2

a) What is the resulting angular speed of the platform? Answer in units of r e v / s .

b)What is the change in kinetic energy of the system? Answer in units of J.

<h3>ANSWER:</h3>

we know that:

Angular Momentum = L = Iω

From conservation of momentum:

Lo = Lf

(Io) (ωo) = (If) (ωf)

ωf = (Io) (ωo)/(If)

ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)

ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec

ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec

The kinetic energy for rotational motion is given as:

K.E = (1/2)Iω²

Thus, the change in kinetic energy will be:

ΔK.E = (K.E)f - (K.E)o

ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²

ΔK.E = (1/2)(Ifωf² - Ioωo²)

ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²

5 0

3 years ago