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1 year ago

Light from a lamp is shining on a surface. how can you increase the intensity of the light on the surface?

1 answer:

4 0

The intensity of a light in a surface follows the inverse square law formula which can be mathematically expressed as,
I = k/d²
where I is intensity, d is distance, and k is the proportionality constant. For us to increase the intensity, we should lower the distance from the source to the surface.

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Two speakers both emit sound of frequency 320 Hz, and are in phase. A receiver sits 2.3 m from one speaker, and 2.9 m from the o

satela [25.4K]

Answer:

Option B

Explanation:

The phase difference is found by subtracting the 2.3m for the receiver from the other speaker which is 2.9m hence

Phase difference= 2.9-2.3= 0.6

3 0

2 years ago

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

6 0

1 year ago

A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction bctwecn thc box and the f

Hitman42 [59]

Answer:

(D) 96 kg-m/s

Explanation:

Let's start off by first calculating the normal force between the box and the floor.

This will be:

Normal Force = 12 * 9.81 = 117.72 N

We can now use the friction equation to find the frictional force on the box when it is moving:

Frictional force = Coefficient of friction * Normal Force

Frictional force = 0.4 * 117.72 = 47.09 N

Finally, since we have the force on the box, we can find the acceleration:

F = Mass * Acceleration

47.09 = 12 * Acceleration

Acceleration = 3.92 m/s^2

Final speed after 2 seconds:

$V=U+a*t$

$V = 4 +(-3.92)*(2)$

V= -3.84 m/s

Since we know the initial and final speeds, we can calculate the change in momentum:

Change in momentum = Final Momentum - Initial Momentum

Change in momentum = $3.84*12-(-4)*12$

Change in momentum = 94.08 kg*m/s

Thus we can see that option (D) is the closest answer.

6 0

2 years ago

The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a

Amanda [17]

Answer:

amount of energy = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and