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2 years ago

A can, containing only air, has its lid tightly screwed on and is left in strong sunlight

Physics

1 answer:

allsm [11]2 years ago

3 0

Answer:

Because everyone knows that when you increase temperature activity withn molecules increases they will collide more making the can probably explode

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A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

v_f = 10.85 m/s

4 0

2 years ago

C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-

mylen [45]

Answer:

C-14 has two more neutrons.

Explanation:

4 0

2 years ago

Read 2 more answers

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

k is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be Q. It is positive.

Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at x = -2.7 m or at x = 0.27 m

3 0

2 years ago

If clouds were made of cotton-candy what is rain

dlinn [17]

Answer:

If clouds were made of cotton candy, I think rain would be soda.

Or liquified sugar.

6 0

1 year ago

Read 2 more answers

PLS HELP QUICK What are the three categories of plant life spans and how long are they?

Temka [501]

Answer:

drag and drop each feature into the correct category

6 0

2 years ago