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Zolol [24]
3 years ago
9

Convert 81.9 km into meters.

Physics
1 answer:
RUDIKE [14]3 years ago
3 0

Answer: 81900

Explanation: 81.9 x 1000 = 81900.

King Henry Doesn’t Usually Drink Chocolate Milk

                                  Meters

(Meters is Usually) (U)

Goes right 3 times so 3 0's so 1000.

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Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.4 m below. The droplets are fal
expeople1 [14]

Answer:

The third drop is 0.26m

Explanation:

The drop 1 impacts at time T is given by:

T=sqrt(2h/g)

T= sqrt[(2×2.4)/9.8]

T= sqrt(4.8/9.8)

T= sqrt(0.4898)

T= 0.70seconds

4th drops starts at dT=0.70/3= 0.23seconds

The interval between the drops is 0.23seconds

Third drop will fall at t= 0.23

h=1/2gt^2

h= 1/2×9.81×(0.23)^2

h= 0.26m

4 0
3 years ago
Mr. Fuller assigned his science class a lab comparing the masses of objects to the nearest gram using a balance. The mass of a p
Tomtit [17]
The scale would need 10 aluminum cubes on one side. Figure out how many paper clips would be needed on the other side to balance this. You have to use more than one aluminum cube because you need to have enough cubes so that you get a whole number mass. 10 cubes gives you a total mass of 27 g for the aluminum.
5 0
3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
If a person weighs 600N on earth, how much does he weigh on the moon
Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0
3 years ago
The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c
Natalka [10]

Answer:

EMF = 1684.67 Volts

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

EMF = \frac{d\phi}{dt}

now we have

\phi = BA

here since magnetic field is constant so we have

EMF = A\frac{dB}{dt}

now we have

A = (190 \times 10^3)(190 \times 10^3)

A = 3.61 \times 10^{10} m^2

now we have

EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})

EMF = 1684.67 Volts

6 0
3 years ago
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