How long would it take light from the sun to reach mercury?

An electron and an antielectron (positron) each have a rest energy of 0.511 MeV , or approximately 8.2×10−14 J . When an electro

Mademuasel [1]

Answer:

Explanation:

Photon is also a particle . Hence when two particles like electron and positron annihilate to get completely changed to photons , a minimum of two photons of equal and opposite momentum and energy are produced flying in opposite direction to conserve momentum and energy . Each photon will have same energy equal to 511 keV . It is so to conserve momentum and energy. Initially total momentum was zero so finally too total momentum should be zero.

8 0

4 years ago

Which material is the best heat insulator? metal wood plastic glass

OlgaM077 [116]

Of the materials listed wood is the best insulator. It would be the least hot if exposed to similar temperatures.

5 0

3 years ago

Read 2 more answers

A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg

Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s. Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0

3 years ago

A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?

kati45 [8]

1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

$v=\frac{S}{t}$

Rearranging the equation, we can write

$S=vt$

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

$S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m$

2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

$S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s$

5 0

4 years ago

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

likoan [24]

Answer:

$U_{b}=+7.3*10^{-8}J$

Explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it $W_{a-b}=-1.9*10^{-8}J$

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

$W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}$

Now substitute the given values

So

$U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J$

3 0

3 years ago