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3 years ago

Calculate 2 x 10^-3cm ÷ 2.5 x 10^4cm

1 answer:

3 0

Here is your answer:

First find the notations:

2×10^-3
=0002
And...
2.5×10^4=25000

Then divide:

0002÷25000=8E-9

Your answer:
=8 x 10-8

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At what point on a roller coaster is the kinetic energy increasing?

xenn [34]

Answer:

Explanation:

5 0

3 years ago

if you transfer equal amounts of geat to 1kg of water and 1kg of cooper, which will change tempature more

Angelina_Jolie [31]

If you transfer equal amounts of geat to 1kg of water and 1kg of cooper, the temperature of the cooper will change more . . . it'll get more degrees warmer than the water will.

That's because the specific geat of water is greater than the specific geat of cooper.

(This just means it takes more geat to warm some mass of water by some amount than it takes to warm the same mass of cooper by the same amount.)

4 0

3 years ago

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is

maksim [4K]

D = v^2 / 2ug

d= 3.5^2 / 0,15 x 9.8 m/s^2

the answer should be around 4.2m

hope this helps

8 0

3 years ago

Leah paid 42.90 for 5 gallons of gas what was the price per gallon of gas leah paid?

oee [108]

Answer:

the price per gallon of gas leah paid is 8.58

Explanation:

42.90/5=8.58

4 0

2 years ago

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity

$v=\sqrt{2.14^2+0.34^2}=2.17m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0$

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0

4 years ago