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2 years ago

What is true of an object pulled inward in an electric field?

Physics

1 answer:

slava [35]2 years ago

5 0

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

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A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this

noname [10]

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

The displacement of the object is d = 3.00 m, while $\theta=180^{\circ}$ , because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

$W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J$

And the work done is negative, because it is done against the motion of the object.

6 0

2 years ago

Read 2 more answers

Consider the vector b⃗ with magnitude 4.00 m at an angle 23.5∘ north of east. what is the x component bx of this vector? express

BlackZzzverrR [31]

Decomposing the vector b on the x-axis and the y-axis, we get a rectangle triangle where the two sides are bx (x-axis) and by (y-axis), and b is the hypothenuse.
The component in x, bx, is equal to the product between the hypothenuse and the cosine of the angle between b and the x-axis, which is $23.5 ^{\circ}$ :
$b_x = b \cos (23^{\circ})=(4.00 m)(\cos (23^{\circ}))=3.68 m$

6 0

2 years ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

The potential energy of a 40 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?

Fynjy0 [20]

The formula for potential energy is PE=mgh
It can have that high of a potential energy if it's relative height what super high.

7 0

2 years ago

When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h

nevsk [136]

Answer:

The correct answer is c. When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

7 0

2 years ago