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3 years ago

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Conceptual question about relativity: Imagine two events--

1 answer:

steposvetlana [31]3 years ago

6 0

They are not the same event in that they occur in different places and times in most frames of reference. In the photon's frame they are not separated in either space nor time because photons don't experience time and at least mathematically all points on the spacetime manifold are the same point to a photon. What the zero spacetime interval can tell us though, is that the events are connected by a light beam (light-like separation). There is as much time between the events as there is space and one event can conceptually cause the other. They are on the cusp between time-like and space-like events.

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If there is an unbalanced force acting on an object, the object will?

Nonamiya [84]

Answer:

A

Explanation:

because newton's second law states that if a resultant force acts on an object then, it will accelerate in the direction of the resultant force

4 0

3 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0

3 years ago

A skater with a mass of 72 kg is traveling east at 5.8 m/s when he collides with another skater of mass 45 kg heading 60° south

Akimi4 [234]

The final velocity is 5.87 m/s

<u>Explanation:</u>

Given-

mass, $m_{1}$ = 72 kg

speed, $v_{1}$ = 5.8 m/s

$Mass_{2}$ , $m_{2}$ = 45 kg

$speed_{2}$ , $v_{2}$ = 12 m/s

Θ = 60°

Final velocity, v = ?

Applying the conservation of momentum:

$m_{1}$ X $v_{1}$ + $m_{2}$ X $v_{2}$ = ( $m_{1}$ + $m_{2}$ ) v

72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v

v = 417.6 + 540 X $\frac{0.5}{117}$

v = 417.6 + $\frac{270}{117}$

v = 5.87 m/s

The final velocity is 5.87 m/s

8 0

3 years ago

The process by which hot and cold air are transferred in the atmosphere is

elena-14-01-66 [18.8K]

Hey there!

<span>The process by which hot and cold air are transferred in the atmosphere is convection.</span>

8 0

3 years ago

Two identical particles of charge 6 μμC and mass 3 μμg are initially at rest and held 3 cm apart. How fast will the particles mo

krek1111 [17]

Answer:

Explanation:

The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .

Their potential energy at distance d

= kq₁q₂ / d

= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J

= 16.2 J

Their total kinetic energy will be equal to this potential energy.

2 x 1/2 x mv² = 16.2

= 3 x 10⁻⁶ v² = 16.2

v = 5.4 x 10⁶

v = 2.32 x 10³ m/s

When masses are different , total P.E, will be divided between them as follows

K E of 3 μ = (16.2 / 30+3) x 30

= 14.73 J

1/2 X 3 X 10⁻⁶ v₁² = 14.73

v₁ = 3.13 x 10³

K E of 30 μ = (16.2 / 30+3) x 3

= 1.47 J

1/2 x 30 x 10⁻⁶ x v₂² = 1.47

v₂ = .313 x 10³ m/s

3 0

3 years ago