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3 years ago

3. Which is not a primary color of light? blue red green yellow

Physics

1 answer:

Margaret [11]3 years ago

5 0

Answer:

Yellow

Explanation:

Since red, green, and blue are primary colors of light, yellow is the only option which is not. So, yellow is not a primary color of light.

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you push a ball to star it rolling along a "perfectly frictionless" surface. How far will the ball roll

-BARSIC- [3]

Technically, it should roll forever.

6 0

3 years ago

Refrigerant-134a enters the expansion valve of a refrigeration system at 160 psia as a saturated liquid and leaves at 30 psia. D

KatRina [158]

Answer:

Temperature : 92.9 F

Internal Energy change: -2.53 Btu/lbm

Explanation:

mh1=mh2

h1=h2

In table A-11 through 13E

p2=120Psi, h1= 41.79 Btu/lbm,

u1=41.49

So T1=90.49 F

P2=20Psi

h2=h1= 41.79 Btu/lbm

T2= -2.43F

u2= 38.96 Btu/lbm

T2-T1 = 92.9 F

u2-u1 = -2.53 Btu/lbm

3 0

3 years ago

6. Directions: Drag each item to the correct location on the diagram. Many of the astronomical events that can be witnessed from

algol13

Rotation- sunset,sunrise,moons movement

all the others are revolution

leave a thanks if i answered your question

4 0

3 years ago

Read 2 more answers

1. Mike wears an 8.50 kg backpack while scaling a cliff. After 90 minutes, Mike is 11.2 m above the starting point.

svet-max [94.6K]

Answer:

a. 952Joules

b. 8008 joules

c. 1.48 watts

8 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago