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deff fn
10 days ago
15

A wheel accelerates uniformly from rest to an angular speed of 25 rad/s in 10 s.(a) Find the angular acceleration of the wheel.

(b) Find the tangential and radial acceleration of a point 10cm from the wheel’s center. (c) How many revolutions has the wheel turned during this time interval? (d) Then, find the wheel’s angular deceleration if it comes to a full stop after 5 rev
Physics
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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
kicyunya [3294]

Response:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Reasoning:

The laser power P = 1 mW = 1 x 10^-3 W

duration t = 250 ms = 250 x 10^-3 s

Taking a beam diameter of 2 mm = 2 x 10^-3 m

therefore

the beam's cross-sectional area A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) The intensity I = P/A

where P refers to the laser's power

and A represents the beam's cross-sectional area

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) The total energy delivered E =Pt

where P is the beam's power

and t is the exposure duration

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field can be computed as

E = \sqrt{2I/ce_{0} }

where I signifies the beam's intensity

and E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9} = 1.55 x 10^-8 v/m

6 0
1 month ago
Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert
Yuliya22 [3333]

To tackle this question, we know the following:

1 Albert equals 88 meters.

1 A = 88 m.

Initially, we square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Thus, one acre is approximately 0.52 square alberts.</span>

7 0
2 months ago
A power cycle operates between a lake’s surface water at a temperature of 300 K and water at a depth whose temperature is 285 K.
inna [3103]

Response: a) 0.04 kW = 40 W

b) 0.05

Explanation:

A)

The thermal efficiency of the power cycle is calculated as Input / Output

Input = 10 kW + 14,400 kJ/min which translates to 10 kW + 14,400 kJ/(60s) = 10 kW + 14,400/60 kW.

Output equals 10 kW

Thus, Thermal Efficiency = Output / Input = 10 kW / 250 kW = 0.04 kW = 40 W

B)

Maximum Thermal Efficiency of the power cycle is defined as 1 - T1/T2

where T1 = 285 Kelvin

and T2 = 300 Kelvin

Thus, Maximum Thermal Efficiency = 1 - T1/T2 = 1 - 285/300 = 0.05

8 0
1 month ago
Read 2 more answers
A student placed a pencil in a cup of water. The pencil appears broken because light- always travels in a straight line makes th
kicyunya [3294]
<span>None of the provided options presents an accurate statement. The third choice is nearly correct,
yet it can be considered misleading.

The pencil seems broken due to light bending away from a straight path
when it transitions from air to water.</span>
5 0
2 months ago
Read 2 more answers
Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi
Keith_Richards [3271]

The answer is:

V=14m/s

Details are as follows:

According to the problem, we have

The combined mass of A and B is 60kg

A's speed is 2m/s

B's speed is 1m/s

The mass of the bag is 5kg

Typically, the momentum of astronaut A along with the bag is defined by

M_A=(60+5)*2

M_A=130kgm/s

To prevent a collision, astronaut A should maintain a speed that is either equal to or less than astronaut B's speed

Thus, the minimum speed astronaut A should achieve corresponds to that of astronaut B, which is 1

Consequently,

130=(60*1)=(5*v)

V=14m/s

7 0
2 months ago
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