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3 years ago

How much energy does a 100 W light bulb use in 24 hours?? Show all your work....

Physics

1 answer:

GalinKa [24]3 years ago

7 0

How much does it use per hour or minute or second? its impossible to answer this question i think

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5. Which of these affect the brightness of a bulb? Choose all that apply.* 6 po

schepotkina [342]

Answer:

The voltage of the battery

7 0

3 years ago

Which would melt first, germanium with a melting point of 1210 k or gold with a melting point of 1064oc?

iren2701 [21]

<span>Germanium To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15 1064 + 273.15 = 1337.15 K So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>

8 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

4 years ago

a 45 kg ice skater initially skating at a velocity of 3 m/s speeds up to a velocity of 5 m/s. calculate the difference in the ma

ad-work [718]

Answer: 90 kgm/s

Explanation:

The momentum (linear momentum) $p$ is given by the following equation:

$p=m.V$

Where:

$m=45 kg$ is the mass of the skater

$V$ is the velocity

In this situation the skater has two values of momentum:

Initial momentum: $p_{1}=m.V_{1}$

Final momentum: $p_{2}=m.V_{2}$

Where:

$V_{1}=3 m/s$

$V_{1}=5 m/s$

So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):

$p=p_{2}-p_{1}=m.V_{2}-m.V_{1}$