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Aleks [24]
3 years ago
5

Which example best illustrates that light behaves like particles?

Physics
2 answers:
AlekseyPX3 years ago
8 0
I would say B. Because actual mass would ricochet off the sidewalk.
mr_godi [17]3 years ago
8 0

Answer:

B. Light bounces off a white cement sidewalk.

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11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh
IrinaVladis [17]

Answer:

The displacement was 320 meters.

Explanation:

Assuming projectile motion and zero initial speed (i.e., the object was dropped, not thrown down), you can calculate the displacement using the kinematic equation:

d = \frac{1}{2}gt^2=\frac{1}{2}10\frac{m}{s^2}\cdot 8^2 s^2=320 m

The displacement was 320 meters.

7 0
3 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

8 0
3 years ago
The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang
Vilka [71]
Normal Force = 54 N
acceleration = 1.2 m/s^2

For Normal Force:
According to the force diagram, we can come up with the equation (all up and down forces):

10 sin 30 + Normal Force - 58.8 = 0
Normal Force = 53.8 N = 54 N

For acceleration:
According to the force diagram, we can come up with the equation (all left and right forces):

10 cos 30 - 1.5 = (6.0) (Acceleration)
Acceleration = 1.19 m/s^2 = 1.2 m/s^2 
3 0
3 years ago
Read 2 more answers
When a object is at rest, what is it’s speed??
nadya68 [22]

Answer:

0   (there is no speed)

Explanation:

If an object is at rest, it is not moving, and it doesn't have a speed, so the speed is zero.

3 0
3 years ago
Read 2 more answers
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