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Marat540 [252]
3 years ago
6

Describe a scenario when you could use energy but not do any work.

Physics
1 answer:
Harlamova29_29 [7]3 years ago
4 0

Answer:

When you don't move, you still use energy. This energy is called potential energy, or, stored energy.

When you don't move or do work, you can use energy.

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figure 2 shows a charged ball of mass m = 1.0 g and charage q = -24*10^-8 c suspended by massless string in the presence of a un
Vlad [161]

Answer:

E = 307667  N/C

Explanation:

Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.

This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:

0.0098 N = T * cos(37)

then T = 0.0098/cos(37) N = 0.01227 N

Knowing the tension's magnitude, we can find its horizontal component:

T * sin(37) = 0.007384 N

and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:

0.007384 N = E * 2.4 * 10^(-8) C

Then  E = 0.007384/2.4 * 10^(-8)  N/C

E = 307667  N/C

8 0
3 years ago
An electric field of intensity 3.7 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.3
jekas [21]

Answer:

a)906.5 Nm^2/C

b) 0

c) 742.56132 N•m^2/C

Explanation:

a) The plane is parallel to the yz-plane.

We know that

flux ∅= EAcosθ

3.7×1000×0.350×0.700=906.5 N•m^2/C

(b) The plane is parallel to the xy-plane.

here theta = 90 degree

therefore,

0  N•m^2/C

(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.

therefore, applying the flux formula we get

3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C

4 0
2 years ago
Read 2 more answers
Applying the Law of Conservation of Energy. If a car was released down the track from a height what happens to the potential ene
erastova [34]

Answer:

According to the law of conservation of energy, energy cannot be created or destroyed,  although it can be changed from one form to another.    KE + PE = constant. A simple example involves a stationary car at the top of a hill.  As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases.  On the way back up the hill, the car converts kinetic energy to potential energy.  In the absence of friction, the car should end up at the same height as it started.

This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum.  When the ball is at the top of its swing, all of the pendulum’s energy is potential energy.   When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy.   The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

4 0
2 years ago
what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s
yaroslaw [1]

Answer : The value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

Explanation :

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant = ?

t = time = 17s

[A_t] = final concentration = 0.0981 M

[A_o] = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}

k=0.51M^{-1}s^{-1}

Therefore, the value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

6 0
3 years ago
In deep space, there is very little friction. Once they launch a probe into deep space, where there are no external forces actin
Charra [1.4K]

Answer:

move at constant velocity.

Explanation:

Newton's first law (also known as law of inertia) states that:

"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".

In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.

5 0
3 years ago
Read 2 more answers
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