Ask question

Ask question

All categories

2 years ago

5

An overhead projector lens is 32.0 cm from a slide (the object) and has a focal length of 30.1 cm. What is the magnification of

the image? PLS HELP

1 answer:

puteri [66]2 years ago

6 0

Answer: 15.8

Explanation:

You are given that the

Object distance U = 32 cm

Focal length F = 30.1 cm

First calculate the image distance V by using the formula

1/F = 1/U + 1/V

Substitute F and V into the formula

1/30.1 = 1/32 + 1/V

1/V = 1/30.1 - 1/32

1/V = 0.00197259

Reciprocate both sides

V = 506.94 cm

Magnification M is the ratio of image distance to object distance.

M = V/U

substitute the values of V and U into the formula

M = 506.94/32

M = 15.8

Therefore, the magnification of the image is 15.8 or approximately 16.

You might be interested in

An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele

leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

$\alpha = \frac{\omega_f - \omega_0}{t}$

Here

$\alpha$ = Angular acceleration

$\omega_{f,i} =$ Initial and final angular velocity

There is not initial angular velocity,then

$\alpha = \frac{\omega_f}{t}$

We know that the relation between the tangential velocity with the angular velocity is given by,

$v = r\omega$

Here,

r = Radius

$\omega$ = Angular velocity,

Rearranging to find the angular velocity

$\omega = \frac{v}{r}}$

$\omega = \frac{30}{0.20} \rightarrow$ Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

$\alpha = \frac{30}{0.20*6}$

$\alpha = 25 rad /s^2$

Therefore teh average angular acceleration of each wheel is $25rad/s^2$

3 0

1 year ago

On her way home from work, Brenda drove 20 miles at 60 miles per hour. Due to poor weather conditions, she then reduced her spee

rjkz [21]

Answer:

D

Explanation:

Speed = distance / time

her time for the first journey = 20 miles / 60 miles/hr = 1/3 hr

her time for second part of the journey = her remaining distance / her speed = (80 - 20) miles / 30 miles/hr = 60 miles / 30 miles/hr = 2 hrs

total time spend by her = 2 hr+ 1/3 hr = 2 1/3 hrs

her traveling the distance at 40 miles per hour = 80 miles / 40 miles /hr = 2 hrs

the time less she would drive if she drive the entire distance at 40 miles/hr = 2 1/3 hrs - 2 hrs = 1/3 hr

3 0

2 years ago

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed

olga55 [171]

Initially, the energies are:

$U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
=K_{i}=\frac{1}{2} m v_{0}^{2}$

At final point, the energies are:

$U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0$

Using conservation law of energy,

$-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}$

The equation is further simplified as:

$r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\
h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\
&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\
& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}$

7 0

1 year ago

3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.

Scrat [10]

Answer:

514 cal

Explanation:

In order to calculate the lost heat by the amount of water you first take into account the following formula:

$Q=mc(T_2-T_1)$ (1)

Q: heat lost by the amount of water = ?

m: mass of the water

c: specific heat of water = 1cal/g°C

T2: final temperature of water = 11°C

T1: initial temperature = 12°C

The amount of water is calculated by using the information about the density of water (1g/ml):

$m=\rho V=(1g/ml)(514ml)=514g$

Then, you replace the values of all parameters in the equation (1):

$Q=(514g)(1cal/g\°C)(11\°C-12\°C)=-514cal$

The amount of water losses a heat of 514 cal

3 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago