All categories

3 years ago

An overhead projector lens is

Physics

1 answer:

Inessa [10]3 years ago

6 0

Answer: 34.9 cm

Explanation:

You are given the following parameters;

Object distance U = 32 cm

Magnification M = - 12.0

According to formula for magnification;

M = V/U

Where V = image distance.

Substitute V and M into the formula

-12 = V/32

Cross multiply

V = -12 × 32

V = - 384

You can use the formula

1/f = 1/V + 1/U

Where f = focal length

Substitute V and U into the formula

1/f = - 1/384 + 1/32

Find the lowest common factor of the denominator at Left hand side

1/f = ( -1 + 12 ) / 384

1/f = 11/384

Reciprocate both sides

F = 384/11

F = 34.9 cm

He should therefore use the focal length of 34.9 cm

You might be interested in

Which of the following describes the effect of a convex mirror on light rays?

natulia [17]

B. It reflects light rays outward.

A convex mirror looks like this ⊂. So when light rays hit the outward curves, it bends the light rays outwards, and backwards, and it would appear to be coming from its center called the focus.

3 0

3 years ago

Refer to the attached image!!!

dimaraw [331]

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

T is time of motion
g is acceleration due to gravity
θ is angle of projection

T = (2 x 12 x sin20)/9.8

T = 0.837 s

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

6 0

2 years ago

A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of

Nesterboy [21]

Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

3 0

2 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago

Canadian Olivia Oliver broke the Guinness World Record for the fastest spin on ice skates on January 2015 with an angular veloci

Shalnov [3]

Answer:

Explanation:

initial angular velocity ω' = 130 rpm

final angular velocity ω = 342 rpm

recall that angular momentum = ωI

where I is the moment of inertia.

for the initial spinning condition, we take moment of inertial = I'

for final spinning condition, we take moment of inertia = I

initial angular momentum = ω'I' = 130 I'

final angular momentum = ωI = 342 I

according to conservation of angular momentum, initial angular momentum must be equal to the final angular momentum, therefore

342 I = 130 I'

ratio of initial moment of inertia to final moment of inertia = I'/I

==> I'/I = 342/130 ≅ <em>2.63 : 1</em>

b) to achieve a final angular velocity of of 375 rpm, her initial velocity will have to be

2.63 = 375/ω'

ω' = 375/2.63 = <em>143.13 rpm</em>

6 0

3 years ago