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kipiarov [429]
2 years ago
13

When an object (like a ball) falls, some of its___ energy changes to ___ energy, due to the law of conservation of energy

Physics
1 answer:
Alinara [238K]2 years ago
8 0

Answer:

potential, kinetic

Explanation:

pls give brainliest :p

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An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp
Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

k=\dfrac{1}{2}mv^2

k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2

k = 1150  J

In two significant figure, k=1.2\times 10^3\ J

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. P=\dfrac{W}{t}=\dfrac{\Delta K}{t}

P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}

P = 115 watts

In two significant figures, P=1.2\times 10^2\ Watts

Hence, this is the required solution.  

6 0
2 years ago
A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas
Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

\varepsilon' =\dfrac{S_y}{E}

\varepsilon' =\dfrac{240}{110\times 1000}

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0
3 years ago
Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a
Novosadov [1.4K]

Answer:

A) correct answer is C,   B)   correct answer is b  and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

        1 / f = 1 / p + 1 / q

where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

        1 / q = 1 / f 1 / p

let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

6 0
3 years ago
Frank does 2400J of work in climbing a set of stairs. If he does the work in 6 seconds, what is his power output?
timurjin [86]
Power = Work Done / Time taken

         Where Work Done is Joules, and Time is in Seconds, Power is in Watts

            = 2400 J / 6 seconds

            = 400 Watts

The power output is 400 Watts.
4 0
3 years ago
What quantity of heat must be removed from 20g<br>of water at 0°C to change it to ice at 0°C?​
seraphim [82]

The quantity of heat must be removed is 1600 cal or 1,6 kcal.

<h3>Explanation : </h3>

From the question we will know if the condition of ice is at the latent point. So, the heat level not affect the temperature, but it can change the object existence. So, for the formula we can use.

\boxed {\bold {Q = m \times L}}

If :

  • Q = heat of latent (cal or J )
  • m = mass of the thing (g or kg)
  • L = latent coefficient (cal/g or J/kg)
<h3>Steps : </h3>

If :

  • m = mass of water = 20 g => its easier if we use kal/g°C
  • L = latent coefficient = 80 cal/g

Q = ... ?

Answer :

Q = m \times L \\ Q = 20 \times 80 = 1600 \: cal

So, the quantity of heat must be removed is 1600 cal or 1,6 kcal.

<u>Subject : Physics </u>

<u>Subject : Physics Keyword : Heat of latent</u>

4 0
3 years ago
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