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3 years ago

1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0

Physics

1 answer:

igor_vitrenko [27]3 years ago

6 0

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0 m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

$\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2$

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

$v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m$

So, the plane slide a distance of 229.5 m.

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The largest salami in the world, made in Norway, was more than 20 m long. If a hungry mouse ran around the salami’s circumferenc

Musya8 [376]

Answer: 0.09 m

Explanation:

Centripetal acceleration in terms of tangential speed is:

$a_c=\frac{v^2_t}{r}$

where r is the radius.

It is given that,

centripetal acceleration of the mouse, $a_c=0.29 m/s^2$

tangential speed , $v_t=0.17m/s$

Radius of salami is:

$r =\frac{(0.17m/s)^2}{0.29 m/s^2}=0.09m$

Thus, the radius of the salami is 0.09 m.

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3 years ago

Which transition by an electron will release the greatest amount of energy? hurry!

Sphinxa [80]

highest energy level to the ground state.

Explanation:

The transition from the highest energy level to the ground state.

An electron has a discrete amount of energy accrued to it in any energy level it belongs to.

Electrons can transition between one energy level or the other.

When electrons change state, they either release or absorb energy.
When an atom absorbs energy, they move from their ground to final state which is consistent with the energy of the final state.
When electrons release energy, they move from excited state to their ground state.
Electrons will release the greatest amount of energy when they move from the highest energy level to the ground state.

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3 years ago

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(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma

andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

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2 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

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3 years ago

Why does the sky change colors at sunset?

KiRa [710]

Atmospheric refraction is the deviation of light or other electromagnetic wave from a straight line as it passes through the atmosphere due to the variation in air density as a function of height. ... Refraction not only affects visible light rays, but all electromagnetic radiation, although in varying degrees.

So in short, the answer is D.

(My answer got deleted because it didnt explain which is dumb)

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2 years ago

Read 2 more answers