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lakkis [162]
3 years ago
5

Someone help with this also

Physics
1 answer:
Mumz [18]3 years ago
3 0
I’m pretty sure it’s average speed= total distance and total time which is A.
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A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start
Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4  m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction =  PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno. 

4 0
3 years ago
The kinetic energy of an object increases as its ____ its potential energy
tankabanditka [31]
The kinetic energy of an object increases as its decreases <span>its potential energy as the sum of energy will remain constant.

In short, Your Answer would be "Decreases"

Hope this helps!</span>
8 0
3 years ago
I need a sentence for the word sandy soil
rusak2 [61]
One day, as I was walking, I found some sandy soil beside the road.
6 0
3 years ago
Read 2 more answers
How long would it take a leopard, running at an average speed of 20 m/s to travel 500 m?
alexandr1967 [171]

Answer:

25 seconds

Explanation:

500/20

4 0
3 years ago
A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0
3 years ago
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