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3 years ago

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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Brittany is changing the tire of her car on a steep hill 22.4 m high. She stumbles and drops the 10.5-kg spare tire, which rolls

iragen [17]

If we assume that there is no friction between the tire and the ground, we can the use formula:
ΔKE + ΔPE = 0
where KE is the kinetic energy which is equal to 0.5mv²
and PE is the potential energy which is equal to mgh

Substituting the given values:
0.5 (10.5) (v2 - 2.10^2) + 10.5 (9.81) (8.59 - 22.4) = 0

Solving for v:
v = 16.59 m/s

5 0

3 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

anygoal [31]

Answer:

$3.95 \times 10^{-14}m$

Explanation:

We are given that

Charged on alpha particle=q=2e= $2\times 1.6\times 10^{-19} C$

Where $e=1.6\times 10^{-19} C$

Initial kinetic energy=K.E=5.76 MeV= $5.76\times 10^6\times 1.6\times 10^{-9} C$

$1 Me V=10^6\times 1.6\times 10^{-19} V$

Z=79

Charge on protons= $q'=79\times 1.6\times 10^{-19} C$

We have to find the closeness of alpha particle to the gold nucleus before being turned around.

Initial kinetic energy=Final potential energy

$5.76\times 10^6\times 1.6\times 10^{-9}=\frac{Kq_1q_2}{r}$

Where $k=9\times 10^9$

$r=\frac{9\times 10^9\times 79\times 1.6\times 10^{-19}\times 2\times 1.6\times 10^{-19}}{5.76\times 10^6\times 1.6\times 10^{-9}}$

$r=3.95\times 10^{-14}m$

7 0

3 years ago

9. The current i in a wire of circular cross section can be calculated from the current density, J, using the following equation

tino4ka555 [31]

Answer:

Explanation:

i = ∫J dA

A = πr²

dA = 2πr dr

i = ∫J 2πr dr

= ∫3 x 10⁸ x 2πr³ dr

=3 x 10⁸ x 2π ∫ r³ dr

integrating and taking limit from r = .9 R to R

3 x 10⁸ x 2π / 4 [ 2⁴ - .9⁴ x 2⁴] x (10⁻³)⁴

= 4.71 x 10⁸ x .3439 x 10⁻¹² x 2⁴

= 25.9 x 10⁻⁴ A

8 0

3 years ago

In schematic diagrams, currents are indicated using arrows. What do the arrows indicate? a) the direction of motion of the elect

netineya [11]

Answer:

Explanation:

The direction that positive charge would move

3 0

3 years ago

A 36,287 kg truck has a momentum of 907,175 kg • . What is the truck’s velocity?

Snowcat [4.5K]

By definition,
Momentum = Mass * Velocity

Let v = the velocity of the truck, m/s
The mass of the truck is 36,287 kg.
The momentum is 907,175 (kg-m)/s.

Therefore
907,175 (kg-m)/s = (36287 kg)*(v m/s)
v = 907175/36287 = 25 m/s

Answer: 25 m/s

7 0

4 years ago

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