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3 years ago

What is the measure of the force that gravity applies to an object

Physics

1 answer:

GarryVolchara [31]3 years ago

4 0

Answer:

Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy-including planets,stars,galaxies, and even light-are brought toward one another.

Explanation:

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An icicle falls off of a skyscraper from rest and falls for 34 seconds. How fast will that icicle be moving after that time? Sho

goblinko [34]

Answer:

The icicle will be moving at 333.54 m/s

Explanation:

Free Fall Motion

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.81 m/s^2$ .

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The icicle falls from rest for 34 seconds. We need to find the speed after that time:

vf = 9.81*34

vf = 333.54 m/s

The icicle will be moving at 333.54 m/s

7 0

3 years ago

A distance between two consecutive troughs in a wave motion train produced in a string is 5 cm. Find speed of wave , if two comp

Alexxx [7]

Speed= (frequency) x (wavelength)

= (2 per second) x (5 cm) = 10 cm/sec .

6 0

4 years ago

A uniform electric field of magnitude 375 n/c pointing in the positive x - direction acts on an electron, which is initially at

Finger [1]

(a) The force exerted by the electric field on the electron is given by the product between the electron charge q and the intensity of the electric field E:
$F=qE=(1.6 \cdot 10^{-19}C)(375 N/C)=6\cdot 10^{-17}N$
Under the action of this force, the electron moves by:
$\Delta x = 3.20 cm=0.032 m$
And the work done by the electric field on the electron is equal to the product between the magnitude of the force and the displacement of the electron. The sign has to be taken as positive, because the direction of the force is the same as the displacement of the electron, so:
$W=F \Delta x= (6\cdot 10^{-17}N)(0.032 m)=1.9 \cdot 10^{-18}J$

(b) The electron is initially at rest and it starts to move under the action of the electric field. This means that as it moves, it acquires kinetic energy and it loses potential energy. The change in potential energy is the opposite of the work done by the electric field:
$\Delta U = U_f - U_i = -1.9 \cdot 10^{-18} J$
Where Uf and Ui are the final and initial potential energy of the electron.

(c) For the conservation of energy, the sum of the kinetic energy and potential energy of the electron at the beginning of the motion and at the end must be equal:
$U_i + K_i = U_f + K_f$ (1)
where Ki and Kf are the initial and final kinetic energies.
The electron is initially at rest, so Ki =0, and we can rewrite (1) as
$U_i - U_f = - \Delta U = K_f = \frac{1}{2}m_e v_f^2$
and by using the mass of the electron me, we can find the value of the final velocity of the electron:
$v_f= \sqrt{ -\frac{2 \Delta U}{m_e} }= \sqrt{- \frac{2(-1.9 \cdot 10^{-18} J)}{9.1 \cdot 10^{-31} kg} } =2.04 \cdot 10^6 m/s$

7 0

3 years ago

What are the types of nuclear radiation ? (Particles let off by nuclear materials.) *

Neko [114]

Answer:

Because nuclear decay releases ionizing radiation (alpha, beta, and gamma radiation) we call such compounds radioactive.

Explanation:

6 0

3 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$