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ivanzaharov [21]
3 years ago
7

A convex mirror, like the passenger-side rearview mirror on a car, has a focal length of -2.2 m . an object is 4.4 m from the mi

rror.
Physics
1 answer:
Maksim231197 [3]3 years ago
4 0
The question is missing, but I guess the problem is asking for the location of the image.

We can solve the problem by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}
where
f is the focal length of the mirror
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

For a convex mirror, the focal length is taken as negative, so in our problem 
f=-2.2 m
The distance the object is
d_o=4.4 m

And if we use the previous equation, we can find the distance of the image from the mirror:
\frac{1}{d_i}= \frac{1}{f}- \frac{1}{d_o}= -\frac{1}{2.2 m} - \frac{1}{4.4 m}    =- \frac{3}{4.4 m}
from which we find
d_i =  -\frac{4.4 m}{3}=-1,47 m
where the negative sign means the image is virtual (located behind the mirror)
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galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N

3 0
3 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
A bullet is fired horizontally at a height of 1.3 meters and a velocity of 950 m/s. How long was the bullet in the air?
seropon [69]

Answer:

<em>The bullet was 0.52 seconds in the air.</em>

Explanation:

<u>Horizontal Motion </u>

It occurs when an object is thrown horizontally with a speed v from a height h.

The object describes a curved path ruled exclusively by gravity until it hits the ground.

To calculate the time the object takes to hit the ground, we use the following equation:

\displaystyle t=\sqrt{\frac{2y}{g}}

Note it doesn't depend on the initial velocity but on the height.

The bullet is fired horizontally at h=1.3 m, thus:

\displaystyle t=\sqrt{\frac{2\cdot 1.3}{9.8}}

\displaystyle t=\sqrt{\frac{2.6}{9.8}}

t = 0.52 s

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3 0
3 years ago
Ecologically speaking, which is bigger, a population or a community? Explain.
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</span>

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If an astronaut can jump straight up to a height of 0.6 m on earth, how high could he jump on the moon?
Zinaida [17]

Answer:

Explanation:

Given

Person on earth can jump to a height s=0.6\ m

initial velocity is u

using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

final velocity is zero

s=\frac{u^2}{2g}

0.6=\frac{u^2}{2g}-----1

On moon surface acceleration due to gravity is \frac{1}{6}[/tex] th of earth gravity

so height attained is given by

h=\frac{u^2\times 6}{2\cdot g}-----2

divide 1 and 2 we get

h=6\times 0.6=3.6\ m                    

4 0
2 years ago
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