Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? A. Use a lens to focus the power into a smaller area. B. Increase the power output of the lamp. C. Either A or B.

Answers

Answer 1

Answer:

Answer:

the correct option is C

Explanation:

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C

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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)

Answers

Answer:

a) W₁ = - 127 J, b) W₂ = 148.18 J, c) $v_(f)$ = 3.43 m/s and d) $v_(f)$ = 3.43 m / s

Explanation:

The work is given the equation

W = F. d

Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them

W = F d cos θ

They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated

W-T = m a

T = W -m a

T = mg -mg/7

T = mg 6/7

T = 3.6 9.8 6/7

T = 30.24 N

Now we can apply the work equation to our problem

a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two

W₁ = F d cos θ

W₁ = 30.24 4.2 cos 180

W₁ = - 127 J

b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)

W₂ = (mg) d cos 0º

W₂ = 3.6 9.8 4.2

W₂ = 148.18 J

c) kinetic energy

K = ½ m v²

Let's calculate speed with kinematics

$v_(f)$ ² = vo² + 2 a y

v₀ = 0

a = g / 7

$v_(f)$ ² = 2g / 7 y

$v_(f)$ = √ (2 9.8 4.2 / 7)

$v_(f)$ = 3.43 m/s

We calculate

K = ½ 3.6 3.43²

K = 21.18 J

d) the speed of the block and we calculate it in the previous part

$v_(f)$ = 3.43 m / s

A 1500 kg car moving at 25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of 2.0 × 106 N/m. What is the maximum distance the spring compresses?

Answers

Answer:

x = 0.68 meters

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Speed of the car, v = 25 m/s

Spring constant of the spring, $k=2* 10^6\ N/m$

When the car hits the uncompressed horizontal ideal spring the kinetic energy of the car is converted to the potential energy of the spring. Let x is the maximum distance compressed by the spring such that,

$(1)/(2)mv^2=(1)/(2)kx^2$

$x=\sqrt{(mv^2)/(k)}$

$x=\sqrt{(1500* (25)^2)/(2* 10^6)}$

x = 0.68 meters

So, the spring is compressed by a distance of 0.68 meters. Hence, this is the required solution.

Final answer:

The maximum distance the spring compresses when a 1500 kg car moving at 25 m/s hits it, given a spring constant of 2.0 × 10⁶N/m, is approximately 0.53 meters or 53 centimeters.

Explanation:

In this specific problem, we can apply the conservation of energy principle, where the initial kinetic energy of the car is converted into potential energy stored in the spring when the car comes to a stop. The formula for kinetic energy is K = 1/2 × m× v² and for potential energy stored in a spring is U = 1/2×k × x², where m = mass of the car, v = velocity of the car, k = spring constant, and x = maximum distance the spring is compressed.

By setting the kinetic energy equal to potential energy (since no energy is lost), we get 1/2 × m×v² = 1/2×k×x². Solving this equation for x (maximum compression of the spring), we obtain x = sqrt((m×v²)/k). Substituting the given values, x = sqrt((1500 kg× (25 m/s)²) / (2.0 × 10⁶ N/m)), which yields approximately 0.53 meters or 53 centimeters. Therefore, the maximum distance the spring compresses is 53 cm.

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A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?

Answers

Answer:

(A). The time is 5.47 sec.

(B). The speed of the rock just before it strikes the ground is 39.59 m/s.

Explanation:

Given that,

Initial velocity = 14.0 m/s

Height = 70.0 m

(A). We need to calculate the time

Using second equation of motion

$s=ut+(1)/(2)gt^2$

Put the value into the formula

$70=-14* t+(1)/(2)*9.8* t^2$

$4.9t^2-14t-70=0$

$t =5.47\ sec$

(B). We need to calculate the speed of the rock just before it strikes the ground

Using third equation of motion

$v^2=u^2+2gs$

Put the value into the formula

$v^2=(14)^2+2*9.8*70$

$v^2=1568$

$v=√(1568)$

$v=39.59\ m/s$

Hence, (A). The time is 5.47 sec.

(B). The speed of the rock just before it strikes the ground is 39.59 m/s.

A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:
d1=_____m
Part B:
d2=______m

Answers

Answer:

Explanation:

In projectile motion , range of projectile is given by the expressions

R = u²sin2θ / g

where u is velocity of projectile.

u = 27 m/s θ = 50

12 = 27² sin 2θ / 9.8

sin 2θ = .16

θ = 9.2 / 2

= 4.6

When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.

Define reflection.what are the two types of reflection

Answers

Answer:

The reflection of light can be roughly categorized into two types of reflection: specular reflection is defined as light reflected from a smooth surface at a definite angle, and diffuse reflection, which is produced by rough surfaces that tend to reflect light in all directions

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Explanation:

Final answer:

Reflection is the process of light bouncing off a surface and changing its direction. There are two types of reflection: specular reflection and diffuse reflection.

Explanation:

Reflection is the process of light bouncing off a surface and changing its direction. There are two types of reflection: specular reflection and diffuse reflection.

Specular reflection occurs when light reflects off a smooth surface, such as a mirror, at a specific angle.

Diffuse reflection occurs when light reflects off a rough surface, such as paper or clothing, and scatters in many different directions.

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determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.

Answers

Answer: i think you should place it on the red line

Explanation:

hope this helps

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