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2 years ago

Is the speed of Earth revolving around the sun changing?

2 answers:

6 0

Answer: First of all, I cannot give you a precise answer, because the speed of the Earth changes all the time as the Earth moves around the Sun. This is because Kepler's second law says that on its orbit, a planet will sweep equal areas in equal amounts of time.

borishaifa [10]2 years ago

5 0

Answer:

30 kilometers per second

Explanation:

the surface of the earth at the equator moves at a speed of 460 meters per second--or roughly 1,000 miles per hour. It covers this route at a speed of nearly 30 kilometers per second, or 67,000 miles per hour.

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A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at

riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0

2 years ago

A ball is thrown horizontally from the top of a building 21.8 m high. The ball strikes the ground at a point 101 m from the base

riadik2000 [5.3K]

Answer:

t=2.10 s

u= 47.40 m/s

Explanation:

given that

h= 21.8 m

x= 101 m

g=9.8 m/s²

Lets take horizontal speed of ball = u m/s

The vertical speed of the car at initial condition is zero ( v= 0).

We know that

$h=vt+\dfrac{1}{2}gt^2$

v= 0 m/s

$h=\dfrac{1}{2}gt^2$

now by putting the values

21.8 = 1/2 x 9.8 x t²

t=2.10 s

This is time when ball was in motion.

Now in horizontal direction

x = u .t

101 = u x 2.1

u= 47.40 m/s

6 0

2 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

Conservation of linear momentum:

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

Conservation of kinetic energy:

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

2 years ago

Which term s the product of force and distance?

goldfiish [28.3K]

Answer:

The answer should be WORK 

(Hope this Helps)

Explanation:

work is equal to the product of force and distance.

W=F*S

W=Work

F=Force applied

S=Distance

8 0

3 years ago

Normally, jet engines push air out the back of the engine, resulting in forward thrust, but commercial aircraft often have thrus

ANEK [815]

Answer:

When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.

Explanation:

Newton’s third law motion states that for every action there will be an equal and opposite reaction.

Thrust reversal is also known as reverse thrust. It acts opposite to the motion of the aircraft by providing the deceleration.

Commercial aircraft moves the ejected air in the forward direction means that the thrust will acts opposite to the motion of the aircraft that is backward direction due to thrust reversal. This thrust force might be used to decelerate the craft.

Uses of thrust reversal in practice:

When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.

6 0

3 years ago