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3 years ago

A 60.watt light bulb runs for 5.0 seconds. How much energy does it use?

Physics

1 answer:

VikaD [51]3 years ago

5 0

"watt" means "Joule of energy per second"

"60 watts" means "60 Joules per second"

(60 joules per second) x (5 seconds) = 300 Joules of energy

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What is most likely to happen to light that hits an opaque object?

ddd [48]

B. is the answer.

C is not correct because the light is actually reflected off of an opaque object.

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3 years ago

What is 0.0025 m^2 in cm^2? 1m = 100cm.

PSYCHO15rus [73]

Answer:

25 cm²

Explanation:

Meters and centimeters are both the units for measuring length. The SI unit of measuring length is meters.

Area is the quantity which measures the cross-section occupied by the object.

Thus,

Given that = Area = 0.0025 m²

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1 m = 100 cm

So, 1 m² = 10000 cm²

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2 years ago

What does keplers second law of planetary monition imply

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Kepler's second law of planetary motion describes the speed of a planet traveling in an elliptical orbit around the sun. It states that a line between the sun and the planetsweeps equal areas in equal times. Thus, the speed of theplanet increases as it nears the sun and decreases as it recedes from the sun.

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3 years ago

Read 2 more answers

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

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2 years ago

Suppose that an electromagnetic wave which is linearly polarized along the x−axis is propagating in vacuum along the z−axis. The

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I think the answer to your question is B

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2 years ago