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Ray Of Light [21]
3 years ago
7

A Roller Derby Exhibition recently came to town. They packed the gym for two

Physics
1 answer:
arlik [135]3 years ago
7 0

Answer:

14.36m/s

Explanation:

From the law of conservation of linear momentum

m1u1 + m2u2 = v(m1 + m2)

68×17 + 76×12= v(68+76)

1156+912 = 144v

2068 = 144v

v = 2068/144

=14.36 m/s

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A 15.7-g aluminum block is warmed to 53.2 °c and plunged into an insulated beaker containing 32.5 g of water initially at 24.5 °
ipn [44]

The equilibrium temperature of aluminium and water is 33.2°C

We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K

Now we can calculate the equilibrium temperature

(mc∆T)_aluminium=(mc∆T)_water

15.7*0.9*(53.2-T)=32.5*1*(T-24.5)

T=33.2°C

6 0
3 years ago
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Who can do my worksheet for me
timurjin [86]

Answer:

what  is it on? like name one of the questions

Explanation:

5 0
3 years ago
Many types of scientific equipment are used to perform different functions in the science lab.Which of the folllowing combinatio
faltersainse [42]

Answer:

Many types of scientific equipment are used to perform different functions in the science lab. Which of the following combinations of equipment would be needed to bring one liter of water to 85°C? a. ... Various pieces of safety equipment are used in the lab to provide protection against injury.

Explanation:

6 0
3 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction
const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

7 0
3 years ago
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