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3 years ago

7

A 26.5-mW laser beam of diameter 1.88 mm is reflected at normal incidence by a perfectly reflecting mirror. Calculate the radiat

ion pressure on the mirror.

1 answer:

Nostrana [21]3 years ago

3 0

Answer:

Explanation:

Area= pier^2

=( 1.88/2)^2= 0.0000000314m^2

Intensity= 0.0265/0.000000314

= 7962W/m^2

Pressure= 2*7962)/345= 5.13*10^-5pa

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A glass ball of radius 3.74 cm sits at the bottom of a container of milk that has a density of 1.04 g/cm3. The normal force on t

Gelneren [198K]

Answer:

The mass of the ball is 0.23 kg

Explanation:

Given that

radius ,r= 3.74 cm

Density of the milk ,ρ = 1.04 g/cm³ = 1.04 x 10⁻³ kg/cm³

Normal force ,N= 9.03 x 10⁻² N

The volume of the ball V

$V=\dfrac{4}{3}\pi r^3$

$V=\dfrac{4}{3}\times \pi \times 3.74^3\ cm^3$

V= 219.13 cm³

The bouncy force on the ball = Fb

Fb = ρ V g

Fb + N = m g

m=Mass of the ball = Density x volume

m = γ V , γ =Density of the Ball

ρ V g + N = γ V g ( take g= 10 m/s²)

$\gamma =\dfrac{N+\rho V g}{V g}$

$\gamma =\dfrac{9.03\times 10^{-2}+1.04\times 10^{-3}\times 219.13\times 10}{219.13\times 10}$

γ = 0.00108 kg/cm³

m = γ V

m = 0.00108 x 219.13

m= 0.23 kg

The mass of the ball is 0.23 kg

5 0

2 years ago

If love is the answer, then what is the question?

kipiarov [429]

Answer:

what is happiness

Explanation:

7 0

3 years ago

Read 2 more answers

A toy rocket is fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the time it wil

miskamm [114]

Answer: 16.3 seconds

Explanation: Given that the

Initial velocity U = 80 ft/s

Let's first calculate the maximum height reached by using third equation of motion.

V^2 = U^2 - 2gH

Where V = final velocity and H = maximum height.

Since the toy is moving against the gravity, g will be negative.

At maximum height, V = 0

0 = 80^2 - 2 × 9.81 × H

6400 = 19.62H

H = 6400/19.62

H = 326.2

Let's us second equation of motion to find time.

H = Ut - 1/2gt^2

Let assume that the ball is dropped from the maximum height. Then,

U = 0. The equation will be reduced to

H = 1/2gt^2

326.2 = 1/2 × 9.81 × t^2

326.2 = 4.905t^2

t^2 = 326.2/4.905

t = sqrt( 66.5 )

t = 8.15 seconds

The time it will take for the rocket to return to ground level will be 2t.

That is, 2 × 8.15 = 16.3 seconds

8 0

3 years ago

A yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth

yulyashka [42]

Answer:

Explanation:

As we know the , equation of time period for simple pendulum ,

T = 2*pi* $\sqrt{l/g}$

hence putting values we get ,

the solution is in picture ,

please

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5 0

3 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago