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jenyasd209 [6]
3 years ago
7

What happened to the combined energy of the two sleds when they collided?

Physics
1 answer:
Komok [63]3 years ago
8 0
Hi!
Here’s your answer:

What happened to the combined energy of the two sleds when they collided? It changed forms into another energy C. Because energy is conserved, the “lost” energy has actually been changed into other forms.

Hope this helps have a good day!!<3
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JulijaS [17]
Total amount of energy would remain constant according to law of conservation of energy. i.e., 50 Joules

In short, Your Answer would be Option C) <span>50 Joules because as energy converts from one form to another, it cannot be created or destroyed during the conversion.
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7 0
3 years ago
Read 2 more answers
In making the determination that bullets, shells, or cartridge cases were fired from a specific weapon, the criminalist would us
ss7ja [257]

Answer:

Comparison Microscope

Explanation:

The Comparison Microscope allows for comparison between two objects or samples by placing them side by side.

It is primarily used in criminology for ballistics which makes it ideal to find out if bullets, shells, or cartridge cases were fired from a specific weapon.

7 0
3 years ago
How does the digestive system help the muscular system?
Sveta_85 [38]

Answer:

I would say the answer is A... but I'm not so sure ....

8 0
3 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

5 0
3 years ago
The gauge pressure at the bottom of a cylinder of liquid is 0.30atm. The liquid is poured into another cylinder with twice the r
likoan [24]

Answer:

P_g' = 0.075 atm

Explanation:

Gauge pressure at the bottom of the cylinder depends on the height of water in the cylinder

So here we can say that

P_g = \rho g h

now when liquid is filled to height "h" in base area "A" then gauge pressure of the liquid at the bottom is given as

P_g = 0.30 atm

now we put the whole liquid into another cylinder with twice radius of the first cylinder

So area becomes 4 times

now by volume conservation we can say that if area is increased by 4 times then height of liquid will decrease by 4 times

so we have

h' = \frac{h}{4}

so gauge pressure is given as

P_g' = \frac{0.30}{4} = 0.075 atm

5 0
3 years ago
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