A 70 Newton dog ran across the yard. What do we know about the dog?

The rhinestones in costume jewelry are glass with index of refraction 1.50. To make them more reflective, they are often coated

Sedaia [141]

Answer:

60

Explanation:

According to the given question, the computation of minimum coating thickness is shown below:-

The condition for constructive interference is

$2t_{min} = (m + \frac{1}{2} )\times \frac{\lambda}{^nmateral}$

$= (0 + \frac{1}{2} )\times \frac{\lambda}{^nmateral}$

$t = \frac{\lambda}{4n}$

Now we will put the values to the above formula to reach the answer

$= \frac{480nm}{4\times 2.0}$

= 60

Therefore we simply applied the above formula to determine the minimum coating thickness

6 0

3 years ago

Kevin goes bowling. Whenever he bowls the ball, he transfers energy from his hand to the bowling ball. The amount of energy befo

g100num [7]

Answer:the answer is C

Explanation:

3 0

3 years ago

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

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

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

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

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

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

c) The peak electric field is given as

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

where I is the intensity of the beam

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

3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago

Distance travelled can be found from the

meriva

Answer:

A

Explanation:

This is because distance traveled (i.e. displacement) is the integral of the velocity function, and velocity is the first derivative of the displacement function. To put this in perspective, the area bounded by a curve can be found by taking the integral of the equation of the curve, taking values on the x-axis as limits.

8 0

3 years ago