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emmasim [6.3K]
2 years ago
14

A convex mirror is placed to the right of an object. The image formed by the mirror will be a

Physics
2 answers:
kotegsom [21]2 years ago
8 0

Answer:If you look at the image of the toy car in the mirror, it will appear to be the same ... However, there is a virtual focal point on the other side of the mirror if we follow them ... Concave mirrors, on the other hand, can have real images. ... Naturally, in concave mirror, the closer the image to the mirror, the bigger the image formed.

svp [43]2 years ago
6 0

Answer:

a

Explanation:

You might be interested in
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,
jonny [76]

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

                                               OR

Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J  

5 0
3 years ago
Why are the parts of an atom that electrons occupy called electron clouds?
Anna [14]

Because it's literally impossible to tell exactly where something that size is
located at any particular time.

And that's NOT because it's so small that we can't see it.  It's because any
material object behaves as if it's made of waves, and the smaller the object is,
the more the size of its waves get to be like the same size as the object. 
When you get down to things the size of subatomic particles, it doesn't make
sense any more to try and talk about where the particle actually "is", and we only
talk about the waves that define it, and how the waves all combine to become a
cloud of <em><u>probability</u></em> of where the particle is.

I know it sounds weird.  But that's the way it is.  Sorry.


8 0
3 years ago
Suppose that the acceleration of a model rocket is proportional to the difference between 100 ft/sec and the rocket's velocity.
sp2606 [1]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 80 ft/s

acceleration, a = 150 ft/s²

Let the time taken is t.

v = u + at

80 = 0 + 150 x t

t = 0.53 second

3 0
3 years ago
If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
2 years ago
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