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3 years ago

7

Which type of energy is commonly referred to as kinetic energy?

2 answers:

nordsb [41]3 years ago

6 0

Answer: C) motion

Explanation: Kinetic energy is the energy possessed by virtue of its motion.

Gravitational energy is the potential energy possessed by a body as virtue of its height.

Mass can contribute to potential. kinetic , gravitational energy.

Wind contributes to wind energy which is used to generate electrical energy.

aksik [14]3 years ago

3 0

Kinetic Energy is movement energy (most simplistic way I can put it) so its motion.

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When light behaves like a particle, it is called a ???

nirvana33 [79]

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4 years ago

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Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

4 years ago

7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If th

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Answer:

Explanation:

Given

mass of children $m_1=20\ kg$

$m_2=30\ kg$

distance between two children $L=3\ m$

suppose small child is at a distance of x m from pivot point

so torque of small child and heavier child must be equal

$20\times (x)=30\times (3-x)$

$2x=9-3x$

$5x=9$

$x=1.8\ m$

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3 years ago

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in the collisio

Dmitriy789 [7]

During the collision between two balls on the pool table there is no external force along the line of collision between them

Since there is no external force on it so here we can say

$F = 0 = \frac{\Delta P}{\Delta t}$

here we have

$\Delta P = 0$

so we can say

$P_i = P_f$

since there is no external force so we can say during the collision the momentum of two balls will remain conserved

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3 years ago

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Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

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4 years ago