PHYSICS COLLEGE

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Answers

Answer 1

Answer:

The correct option is D

Explanation:

From the question we are told that

The maximum electric field strength is $E = 1.0 *10^(11) \ V/m$

The area is $A = 5.0 \ mm^2 = 5.0 *10^(-6) \ m^2$

Generally the force the laser applies is mathematically represented as

$F = \epsilon_o * E ^2 * A$

Here $\epsilon_o = 8.85*10^(-12) C/(V \cdot m)$

$F = 8.85*10^(-12) * (1.0 *10^(11)) ^2 * 5.00*10^(-6 )$

=> $F = 4.43 *10^(5) \ N$

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Talia is on a road trip with some friends. In the first 2 hours, they travel 100 miles. Then they hit traffic and go only 30 miles in the next hour. The last hour of their trip, they drive 75 miles.Calculate the average speed of Talia’s car during the trip. Give your answer to the nearest whole number.
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A 1.7 kg model airplane is flying north at 12.5 m/s initially, and 25 seconds later is observed heading 30 degrees west of north at 25 m/s. What is the magnitude of the average net force on the airplane during this time interval?

A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil has 100 turns how much current must pass through the coil?

Answers

Answer:

The current pass through the coil is 6.25 A

Explanation:

Given that,

Diameter = 25 cm

Magnetic field = 1.0 mT

Number of turns = 100

We need to calculate the current

Using the formula of magnetic field

$B =(\mu_(0)NI)/(2\pi r)$

$I=(B*2\pi r)/(\mu N)$

Where, N = number of turns

r = radius

I = current

Put the value into the formula

$I=(1.0*10^(-3)*2\pi*12.5*10^(-2))/(4\pi*10^(-7)100)$

$I=6.25\ A$

Hence, The current passes through the coil is 6.25 A

According to the World Flying Disk Federation, the world distance record for a flying disk throw in the men’s 85-years-and-older category is held by Jack Roddick of Pennsylvania, who on July 13, 2007, at the age of 86, threw a flying disk for a distance of 54.0 m. If the flying disk was thrown horizontally with a speed of 13.0 m/s, how long did the flying disk remain aloft? (Jack Roddick was also a physics teacher! Read more about him at

Answers

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

$t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s$

Hence, for 4.15 seconds the flying disk remain aloft.

The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.

Answers

Answer:

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Explanation:

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The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. 1.11 × 104 Pa 1.09 × 105 Pa 1.13 × 107 Pa 1.11 × 108 Pa 2.18 × 105 Pa

Answers

Answer: 1.11 x 10⁸ Pa

Explanation:

At any deep, the absolute pressure is the same for all points located at the same level, and can be expressed as follows:

p = p₀ + δ. g . h, where p₀ = atmospheric pressure = 101, 325 Pa

Replacing by the values, we get:

p= 101,325 Pa + 1025 Kg/m³ . 9.8 m/s². 11,033 m = 1.11 x 10⁸ Pa.

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answers

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = (92)/(9.8) = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_(net) = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_(net) = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_(net) = \Delta Kinetic \ Energy$

$W_(net) = (1)/(2) m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_(net) = (1)/(2) * m v^2$

Making v the subject

$v = \sqrt{(2 W_(net))/(m) }$

Substituting value

$v = \sqrt{(2 * 309.98)/(9.286) }$

$v =8.17 m/s$

A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation