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lys-0071 [83]
3 years ago
12

In which of the following does enstein's famous equation apply?

Physics
1 answer:
Delicious77 [7]3 years ago
7 0

Answer:

D

Explanation:

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A car with a mass of 725 kg is moving at 32 m/s toward the east. What is the momentum of the car? A second car with a mass of 2,
Elden [556K]

Answer:

A compact car, with mass 725 kg, ... at 115 km/h toward the east. ... b. A second car, with a mass of 2175 kg, has the same momentum. What is its ... Glisens : m = 2175 kg;. D 21,32 xrolka. anknown. r = ? 110.6 mis east

Explanation:

7 0
3 years ago
Mike walks 200 km in 6 hours.he then walks another 100km in 4 hours .what is his average speed?
kolbaska11 [484]

Average speed is defined as the ratio of total distance covered in total given time

speed = \frac{distance}{time}

here we know that total distance that man moved is

d_1 = 200 km

d_2 = 100 km

so total distance is

d = d_1 + d_2

d = 200 km + 100 km

d = 300 km

now here total time of the motion is

t_1 = 6 hours

t_2 = 4 hours

total time will be given as

t = t_1 + t_2

t = 6 + 4 = 10 hours

now by above formula

v_{avg} = \frac{300}{10}

v_{avg} = 30 km/h

so his average speed is 30 km/h

8 0
3 years ago
Please answer correctly to be marked as brainly.
lidiya [134]

A- PRICE

B-QUANTITY

C-SUPPLY

D-DEMAND

E-EQUILIBRIUM POINT

Explanation:

It is the Supply Demand curve in Economics. It gives relationship between price and quantity

7 0
3 years ago
Chemical properties of an acid include...
Cerrena [4.2K]

and C. corrosive, increases the concentration of hydrogen ions when added to water, forms hydrogen gas when it comes in contact with a metal, and formssalt and water when added to a base.

5 0
3 years ago
Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
2aS=v_f^2 - v_i^2
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
v_f =0
So we can rewrite the equation as
2(-9.81 m/s^2) h=-v_i^2
from which we can isolate h
h= \frac{v_i^2}{19.62} (1)

Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
8 0
3 years ago
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