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3 years ago

What is the difference between potential and kinetic types of energy? 1 pois

2 answers:

5 0

Answer: The main difference between potential and kinetic energy is that one is the energy of what can be and one is the energy of what is. In other words, potential energy is stationary, with stored energy to be released; kinetic energy is energy in motion, actively using energy for movement.
Explanation: hoped the helped :)

Tasya [4]3 years ago

3 0

Answer:

Potential energy is energy that is stored in an object or system. It remains unaffected by the environment outside of the object or system. Kinetic energy is the energy of an object or a system's particles in motion.

Explanation:

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Draw an energy chain diagram to show energy transformations for this event:

larisa86 [58]

The energy chain diagram in this case is chemical energy >> energy conversion >> motion (kinetic) energy + thermal energy.

<h3>What is an energy chain diagram?</h3>

An energy chain diagram is a graphic representation indicating the conversion between different types of energies.

Kinetic energy is a type of motion (movement) energy generated by using potential (stored) energy.

Chemical energy (in this case, the fuel of the car) is a type of energy that is stored to perform work.

Learn more about kinetic energy here:

brainly.com/question/25959744

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2 years ago

" A bowl of soup at 200Á F. is placed in a room of constant temperature of 60Á F. The

Dahasolnce [82]

T(t)=60+140e−0.075t T(12)=60+140e−0.075∗12 T(12)=60+140e−0.9 T(12)=60+140(0.4065696597)
=116.84
So the temperature will be approximately 117 degrees

7 0

3 years ago

A hockey puck has a coefficient of kinetic friction of μk = .35. If the puck feels a normal force (FN) of 5 N, what is the frict

alina1380 [7]

Answer:

The frictional force is $F_f = 1.75 \ N$

Explanation:

From the question we are told that

The coefficient of kinetic force is μk = 0.35

The normal force felt by the puck is $F_N = 5 \ N$

Generally the frictional force that acts on the puck is mathematically represented as

$F_f = \mu_k * F_N$

=> $F_f = 0.35 * 5$

=> $F_f = 1.75 \ N$

3 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

A clock moves past you at a speed of 0.9 c. How much time passes for you for each second that elapses on the moving clock?

Zigmanuir [339]

Answer:

Explanation:

Time dilation formula is

T = T₀ / √ 1-v²/c²

T₀ is time elapsed in moving reference , T time elapsed in stationary reference.

Here T₀ = 1 second

T = 1/√ 1-0.9² = 1/.4358 = 2.3 second

So 2.3 second will pass for each second on moving reference.

6 0

3 years ago

What is the difference between potential and kinetic types of energy? 1 pois​

What is the difference between potential and kinetic types of energy? 1 pois