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3 years ago

What is the weight of a feather (mass = 0.0001 kg) that floats through earth's and the moon's atmospheres?

Physics

1 answer:

olya-2409 [2.1K]3 years ago

6 0

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth: (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>

On the Moon: (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>

The presence or absence of atmosphere makes no difference. In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup. Weight is just the force of gravity between the feather and the Earth. It's not affected by what's around the feather, or what's happening to it.

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3 years ago

Light of a given wavelength is used to illuminate the surfacce of a metal, however, no photoelectrons are emitted. In order to c

Jlenok [28]

Answer:

B. use light of a shorter wavelength.

Explanation:

We know that

$E= \frac{hc}{\lambda}$

h= plank's constant

c= speed of light

λ= wavelength of the incident light

so, in order to have sufficient energy for for the emission of electron, the incident's radiation must have λ small enough.

B. use light of a shorter wavelength.

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3 years ago

the kinetic energy of an object with mass m moving with a velocity of 5 m/s is 25 j what will be its kinetic energy when its vel

Dmitriy789 [7]

Answer:

100 J, 225 J

Explanation:

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the velocity of the object

In this problem, the initial kinetic energy of the object is

K = 25 J

Then, the velocity is doubled, which means

v' = 2v

Therefore, the new kinetic energy will be

$K'=\frac{1}{2}m(2v)^2 = 4(\frac{1}{2}mv^2)=4K$

Therefore, the kinetic energy has quadrupled:

$K' = 4(25)=100 J$

Later, the velocity is tripled, which means

v'' = 3v

Therefore, the new kinetic energy will be

$K''=\frac{1}{2}m(3v)^2 = 9(\frac{1}{2}mv^2)=9K$

Therefore, the kinetic energy has increased by a factor of 9:

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3 years ago

A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu

seropon [69]

Answer:

$\theta=50\ revolution$

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, $\omega_i=600\ rpm=62.83\ rad/s$

Finally it shuts off, $\omega_f=0$

Time taken, t = 10 s

Let $\alpha$ is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{0-62.83}{10}$

$\alpha =-6.283\ rad/s^2$

Let $\theta$ is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }$

$\theta=\dfrac{-62.83^2}{2\times -6.283}$

$\theta=314.15\ radian$

$\theta=49.99\ revolution$

$\theta=50\ revolution$

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0

3 years ago

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

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third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

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3 years ago