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3 years ago

If a light source (such as a light bulb) appears white, then it must _________ all colors of light.

Physics

1 answer:

pishuonlain [190]3 years ago

5 0

Emit means put out so, emit

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A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.

konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

$W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J$

Learn more about work and kinetic energy:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/6536722

#LearnwithBrainly

6 0

3 years ago

Every football field has two 300kg field goal posts separated by 110m of football field (that includes the endzones). What is th

Mama L [17]

Answer:

4.96×10¯¹⁰ N

Explanation:

The following data were obtained from the question:

Mass 1 (M1) = 300 Kg

Mass 2 (M2) = 300 Kg

Separating distance (r) = 110 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Gravitational force (F) =?

The gravitational force between the two goal posts can be obtained as follow:

F = GM1M2 / r²

F = 6.67×10¯¹¹ × 300 × 300 / 110²

F = 6.003×10¯⁶ / 12100

F = 4.96×10¯¹⁰ N

Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N

3 0

2 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago

Why does the sound of a person’s voice sound different underwater than when they are speaking in air?

Tems11 [23]

Sound waves actually travel much faster in water than air, but words and the direction of the noise are distorted.

3 0

3 years ago

A car with mass 1500 kg moves with constant velocity of 36 m/s. The driver sees a group of cows in front and he immediately step

Crazy boy [7]

From laws of motion:

$S = ( \frac{v + u}{2} ) \times t$
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s

Substitute the values, hence:
$S = ( \frac{0 + 36}{2} ) \times 6$
$S = (18) \times 6 \\ \\ S = 108m$

But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.

Therefore, the distance between the car and the cows = 160-108
Distance = 52m

6 0

3 years ago