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DanielleElmas [232]
3 years ago
7

Thermal convection applies mainly to _________.

Physics
2 answers:
Serggg [28]3 years ago
8 0

That's the circulation of fluids on account of different densities at different places in the fluid, which almost always means different temperatures.

Basile [38]3 years ago
7 0

the answer should be "fluids".

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Which organisms are the oldest to most recent, based on scientific views?
amid [387]

go with D. cyanobactweia,sea scorpion,birds,saber-toothed cats
5 0
3 years ago
Which particles account for most of the mass of an atom?
BartSMP [9]
Protons and Neutrons account for most of an atoms mass
8 0
3 years ago
An astronaut drops a rock from the top of a crater on the moon. When the rock is halfway down to the bottom of the crater, its s
Alexxx [7]

Answer: vf1/vf2= 1/ sqrt(2)

Explanation :on the moon no drag force so we have only the  force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics

if the rock travels H to the bottom we can calculate velocity:

vo=0m/s (drops the rock)  , yo=0

vf*vf= vo*vo+2g(y-yo)

when the rock is halfway  y = H/2 so:

vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)

when the rock reach the bottom y=H so:

vf2*vf2=2*g*H so vf2 = sqrt(2gH)

so vf1/vf2= 1/ sqrt(2)

good luck from colombia

8 0
3 years ago
Read 2 more answers
Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com
Bas_tet [7]

Answer:

Failure rate   = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at =  140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷  (number of breakdowns)    ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be  = (Total up time) ÷ (number of breakdowns)      .......2

=  \frac{2000}{2}   =  1000  

so here gap between occurrences is

gap between occurrences=  140 - 20

gap between occurrences = 120 hour

and

MTBF  will be

MTBF = 1000 - 120

MTBF = 880 hours  

and

Failure rate (FR)  will be

Failure rate (FR) =  1 ÷ MTBF    ................3

Failure rate (FR) = R÷T     ......................4

as here R is the number of failures and T is total time

so Failure rate (FR)  = 20%

4 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!!!

4 0
3 years ago
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