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In-s [12.5K]
3 years ago
14

How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example

Physics
1 answer:
MariettaO [177]3 years ago
4 0

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

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What effect or effects would be most significant if the Moon's orbital plane were exactly the same as the ecliptic plane?
Lilit [14]

Answer:

c. Solar eclipses would be much more frequent.

Explanation:

The <u>ecliptic plane</u> is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.

A <u>solar eclipse</u> occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.

The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.

So <u>if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.</u>

So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.

the answer is: c.

8 0
4 years ago
A basketball player does 2.43 x 105 J of work during her time in the game, and evaporates 0,1 '10 kg of water. Assuming a latent
olchik [2.2K]

The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal

For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.

The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.

From the given information,

  • the work done on the basketball is dW = 2.43 × 10⁵ J

The amount of heat loss is represented by dQ.

where;

  • dQ = -mL

∴

Using the first law of thermodynamics:b

dU = dQ - dW

dU = -mL - dW

dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)

dU = -491.6 × 10³ J

dU = -491.6 kJ

The number of nutritional calories the player has converted to work and heat can be determined by using the relation:

\mathbf{dU = -491.6 \ kJ \times (\dfrac{1 \ cal}{ 4.186 \ J})}

dU = -117.44 kcal

Learn more about first law of thermodynamics here:

brainly.com/question/3808473?referrer=searchResults

5 0
2 years ago
Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli
Masja [62]

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0
3 years ago
Find the area under the standard normal curve to the right of z=0.49. round your answer to four decimal places, if necessary.
VikaD [51]

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

<h3>What is  the standard normal curve?</h3>

The horizontal axis is approached by the standard normal bend as it extends indefinitely both in directions without ever being touched by it. The center of the bell-shaped, z=0 standard normal curve. Between z=3 and z=3, almost the entire area underneath the standard normal curve is located.

<h3>Use of the standard normal curve:</h3>

Use the normal distribution's standard form to calculate probability. Since the standard normal distribution is indeed a probability distribution, the probability that a variable will take on a range of values is indicated by area of the curve between two points. 100% or 1 is the total area beneath the curve.

<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

= P[z < 0.49] + 1 -  P[ z < 2.05]

= .6879 + 1 - .9798

= 0.7081

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

To know more about standard normal curve visit:

brainly.com/question/12972781

#SPJ4

I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

4 0
2 years ago
A pitcher can throw a fastball that reaches home plate at 95 mph. What is this speed in m/s
Taya2010 [7]
42.47 meters per second
5 0
3 years ago
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