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wlad13 [49]
3 years ago
13

An object is placed 110 mm from a concave mirror with a focal length of 5.0 cm.

Physics
1 answer:
Nady [450]3 years ago
7 0

Answer:

Approximately 92 mm.

Explanation:

Object's distance, u = 110 mm = 11 cm

focal length of the mirror = 5 cm

Image distance, v = ?

To determine the image distance, apply the mirror's formula;

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

   = \frac{1}{5} - \frac{1}{11}

   = \frac{11-5}{55}

\frac{1}{v} = \frac{6}{55}

⇒ v = \frac{55}{6}

       = 9.166667

v = 9.2 cm

The image distance is 92 mm.

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Why does air near the equator rise
Vesnalui [34]
Because it’s warmer and more humid
8 0
4 years ago
Starting from rest, an intern pushes a 42-kggurney 12 m down the hall with a constant force of 80 N directed downward at an angl
DerKrebs [107]

Answer:

A. The work done by the intern is 792 J.

B. The velocity of the gurney when it has moved 12 m is 6.1 m/s.

C. The 12-m journey takes 3.8 s.

Explanation:

Hi there!

Please see the attached figure for a description of the situation.

<u>Part A: </u>

Work is done by a force when it is applied in the direction of the displacement or against it. In this case, the only force applied in the direction of displacement is the horizontal component of the force applied by the intern.

By trigonometry, the horizontal component of the force is calculated as follows:

cos θ = adjacent/hypotenuse

Looking at the figure, you can notice that the applied force, F, is the hypotenuse of a right triangle and the horizontal component, Fx, is the adjacent side:

cos θ = Fx / F  

Fx = F · cos θ

Fx = 80 N · cos 35°

Fx = 66 N

Now we can calculate the work (W) done by this force:

W = Fx · x

Where x is the displacement:

W = 66 N · 12 m = 792 J

The work done by the intern is 792 J.

<u>Part B:</u>

Applying the work-energy theorem, the work done is equal to the change in kinetic energy:

W = final kinetic energy - initial kinetic energy

Since the gurney starts from rest, the initial kinetic energy is zero. Then:

W = final kinetic energy

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the gurney.

v = velocity.

KE = 792 J

792 J = 1/2 · 42 kg · v²

v²= 2· 792 J / 42 kg

v = 6.1 m/s

The velocity of the gurney when it has moved 12 m is 6.1 m/s.

<u>Part C:</u>

First, let´s find the acceleration of the gurney:

Fx = m · a

Fx/m = a

66N / 42 kg = a

a = 1.6 m/s²

Now using the equation of velocity, let´s find the time at which the gurney has a velocity of 6.1 m/s:

v = v0 + a · t

Where:

v = velocity at time t.

v0 = initial velocity.

a = accleration.

t = time.

v = v0 + a · t

6.1 m/s = 0 + 1.6 m/s² · t

t = 6.1 m/s / 1.6 m/s²

t = 3.8 s

The 12-m journey takes 3.8 s.

3 0
3 years ago
A point charge has q=1.0×10-6
dmitriy555 [2]

Since the 2 points form a triangle with hypothenuse of √5 [ √(1²+2²)],

I guess apply the formula :

v = kq \div r

with r as √5 and q as 1x10^-6

not sure about this answer tho

draw a diagram first to understand better

8 0
3 years ago
An object is thrown vertically upward with a velocity of 14m/s for 1.43 second to reach the highest point find the distance cove
Hitman42 [59]

Please find attached photograph for your answer.

4 0
3 years ago
A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0
4 years ago
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