All categories

3 years ago

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

Physics

1 answer:

slava [35]3 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

You might be interested in

9. [03.03]

Evgen [1.6K]

Answer:

Circuit one will have more current than circuit two

Explanation:

I am assuming that you have to see which circuit has the greater current in this case. Well, this is the perfect example of Ohm's Law, which states the following -

V = IR,

where V = voltage / potential difference, I = current, and R = resistance

If one circuit has twice the voltage and half the resistance of the second circuit, as voltage is directly proportional to the resistance -

2V = I( 1 / 2R ),

4V = IR,

I = 4V / R

Whereas in the second circuit -

V = IR,

I = V / R

As you can note, voltage is directly proportional to the current ( I ) as well as the resistance. The only difference between the two formulas I = 4V / R, and I = V / R is the difference in the voltage. With the voltage being 4 times greater in the first circuit, and current is 4 times greater in the first circuit as well.

<u><em>Hence, circuit one will have more current than circuit two</em></u>

5 0

3 years ago

Read 2 more answers

You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the

Vinvika [58]

Answer:

Weight of the car, normal force, drag force

Explanation:

The forces acting on the car are:

The normal force which acts perpendicularly to the downhill plane
The weight of the car which acts vertically downwards
The drag force due to air resistance which acts in opposition to the motion of the car

Friction is ignored, so the force due to friction is assumed negligible

6 0

3 years ago

A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine t

LuckyWell [14K]

Answer:

Tension of the wire(T) = 169 N

Explanation:

Given:

f = 65Hz

Length of the piano wire (L) = 2 m

Mass density = 5.0 g/m² = 0.005 kg/m²

Find:

Tension of the wire(T)

Computation:

f = v / λ

65 = v / 2L

65 = v /(2)(2)

v = 260 m/s

T = v² (m/l)

T = (260)²(0.005/2)

T = 169 N

Tension of the wire(T) = 169 N

6 0

3 years ago

A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.

NARA [144]

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:

- A uniform motion with constant velocity along the horizontal direction

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

$u_y=0$ is the initial vertical velocity (the missile is thrown horizontally)

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s$

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

$v_x = 45 m/s$

Therefore, the distance covered in a time t is

$d=v_x t$

and by substituting t = 3.91 s, we find the horizontal range of the missile:

$d=(45)(3.91)=176 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago

Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the

lesya [120]

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

$F_{net}$ = - $W_x$ - $f_s$

The $W_x$ is a x-component of force due to gravity (W) and, in this case, is given by: $W_x$ = W.sin(14)

W is described as: W = m.g

Force due to friction ( $f_s$ ) is given by: $f_s$ = μs.N

N is the normal force and, in the system, is equivalent of $W_y$ , so:

$W_y$ = m.g.cos(14)

Therefore, the formula will be:

$F_{net}$ = - $W_x$ - $f_s$

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

3 0

3 years ago