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2 years ago

Match the following examples of energy with the primary form of energy exhibited.

Physics

2 answers:

Travka [436]2 years ago

8 0

Answer:

• Friction ( heat )

• Nuclear power plant ( nuclear )

• Toaster element ( heat )

• welding torch ( heat )

• eraser sitting on a desk edge (potential)

• light bulb ( light )

• campfire ( heat )

• moving car ( kinetic )

• lump of coal in a storage bin ( potential )

• stick of tent ( potential)

Explanation: hope this helps !

GalinKa [24]2 years ago

3 0

The primary energy exhibited is:

friction: Heat energy

Due to friction. there is heat loss in mostly all machines.

nuclear power plant: Nuclear energy

In nuclear power plant, fission or fusion of radio-active substances produces large amount of nuclear energy.

toaster element: Heat energy

Heat energy is produced in the toaster when electricity passes through the element. This is used to heat the toasts.

welding torch: Heat energy

The heat energy produced in the welding torch is used in welding.

eraser sitting on a desk edge: Potential energy

Potential energy is possessed by a body due to virtue of its height.

light bulb: Heat energy and light energy.

The element inside the bulb gets heated first and then light energy is produced.

campfire: Heat energy

Heat is produced and then light energy in case of campfire.

moving car: Kinetic energy

A moving body has kinetic energy.

lump of coal in a storage bin: Potential energy

A lump of coal in a storage bin has potential energy.

stick of TNT: potential energy

Stick of TNT is formed of explosive mixture. It contains potential energy.

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Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.

trapecia [35]

The formula used to find potential energy is P.E. = M * G * H (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is 5 * 9.8 * 2, which equals 98.

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3 years ago

Young's Modulus refers to changes in the: a- Volume b- Length c- Body layers

bekas [8.4K]

Answer: The correct answer is Option b.

Explanation:

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

Stress is defined as force per unit area and strain is defined as proportional deformation in a material.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight

l = length of wire

A = area of cross section

$\Delta l$ = change in length

Hence, the correct answer is Option b.

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3 years ago

If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, its t

Lina20 [59]

$E_{total} = KE + PE \\ KE = 500 J \\ PE = 1500 J \\ E_{total} = 500J + 1500J \\ E_{total} = 2000J$

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3 years ago

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How much time will it take for a coyote to travel 48 meters across the field to get to the unsuspecting rabbit eating grass in t

kkurt [141]

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t=d/s
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t=12s

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3 years ago

A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma

goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

$F=ma$

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2$

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

$F=ma$

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2$

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

$x(t)=\frac{1}{2}a_g t^2$

where $a_g = 0.098 m/s^2$ is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

$x'(t)=15-\frac{1}{2}a_s t^2$

where $a_s=0.8 m/s^2$ is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

$\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s$

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3 years ago