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kirill [66]
3 years ago
13

The total yearly world consumption of energy is approximately 4.0 × 1020 J. How much mass would have to be completely converted

into energy to provide this amount of energy?
Physics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

m = 4.4 × 10³ kg

Explanation:

Given that:

The total yearly energy is 4.0 × 10²⁰ J

The amount of mass that provides this energy can be determined by using the formula:

E = mc²

where;

c = speed of light in free space = (3 × 10⁸)

4.0 × 10²⁰ = m × (3 × 10⁸)²

m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}

m = 4.4 × 10³ kg

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Abnormal protrusion of the eye out of the orbit is known as
kifflom [539]

Answer:

Exophthalmos

Explanation:

Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.

It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.

It is commonly caused by trauma or swelling of eye surrounding tissues resulting from trauma.

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3 years ago
Plate tectonics is the scientific theory explaining the movement of the earths crust. The movement of these tectonic plates is l
zzz [600]

Answer:

Continental drift theory describes the long term effect of plate tectonics.

Explanation:

The long term result of plate tectonic movement is the continental drift. The continents of Earth lay on tectonic plates, that are in motion and interaction via plate tectonics. The drift of the Earths continent is an ongoing process evident in the rift valleys and seafloor spreading zones.

The theory that  the Earth's continents are dynamic and have drifted relative to each other is known as continental drift which correlates with the theory of plate tectonics.

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7 0
3 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

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garik1379 [7]
Can you please stop pasting this question, just go to his profile and ask him.
7 0
3 years ago
A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro
NARA [144]

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width  = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

5 0
3 years ago
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