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Mariulka [41]
3 years ago
14

Paul is driving his car on a sunny afternoon. Which glasses should he ideally wear while driving?

Physics
1 answer:
earnstyle [38]3 years ago
6 0

Answer:

Polarized sunglasses

Explanation:

Polarized sunglasses are comprised of polarized lenses which have a special  type of filter that can block the strong reflected light emitted from the sun. It also helps in decreasing the beam of light and eventually reduces the discomfort during the bright sunlight. It reduces the eye strain to see the objects clearly and visibly.

These sunglasses are mostly used while driving and protects our eyes from the harmful incoming solar lights and enables a person to observe the path clearly without any disturbance.

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a ball is thrown horizontally from a 20 m high building with a speed of 5.0 m/s. How far from the base of the building does the
kipiarov [429]
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2

Find:
Horizontal displacement

Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s

Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
7 0
3 years ago
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
emmainna [20.7K]

Answer:

5,2

Explanation:

From the question we are told that:

Speed of light C=3.0×10^8 m/s.

Generally the equation for Average Speed is mathematically given by

V_{avg}=\frac{d}{t}

Where

d=Distance between the Earth and the sun

d=1.5*10^11m

Therefore

t=\frac{d}{V_{avg}}

t=\frac{1.5*10^11m}{3.0×10^8 m/s.}

t=5*10^2s

Since m and n is given in the form of

m*10^n

Therefore

m=5 & n=2

5,2

3 0
3 years ago
When do sunspots disappear?
dedylja [7]

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

once the magnetic field weakens the area and cold plasma enters the area of the sunspot

8 0
2 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
Converting 15 miles to kilometer
Dmitry [639]
15 miles to kilometers would be: 24.14 kilometers
8 0
3 years ago
Read 2 more answers
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