I think there's a typo because the answer I'm getting is very large.
This is what I'm getting
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c = speed of light
c = 3.0 x 10^8 m/sec approximately
This is roughly 300 million meters per second
The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.
distance = rate*time
d = r*t
d = (3.0 x 10^8) * (7.0 x 10^2)
d = (3.0*7.0) x (10^8*10^2)
d = 21.0 x 10^(8+2)
d = 21.0 x 10^10
d = (2.1 x 10^1) * 10^10
d = 2.1 x (10^1*10^10)
d = 2.1 x 10^11 meters
d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)
Answer:
The time for the cake to cool off to room temperature is
approximately 30 minutes.
Let
=
F be the temperature and T that of the body
Explanation:
Our Tm = 70, the initial-value problem is
= <em>k</em>(T − 70), T(0) = 300
Solving the equation, we get
= <em>kdt</em>
In [T-70]= <em>kt </em>+
T = 70 +

Finding he value for
using the initial value of T (0)= 300, therefore we get:
300=70+
= 230 therefore
T= 70+ 230 
Finding the value for <em>k </em>using T (3) = 200, therefore we get
T (3) = 200
= 
<em>K </em>=
in 
= -0.19018
Therefore
T(t) = 70+230
Answer:
Explanation:
a)
Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.
100
x 10cm = 1000
Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....
x 1000
= 1000g or 1kg
Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....
1000g + 100g = 1100g or 1.1kg
b)
The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....
20g ÷
= 2.5 
c)
Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5
overflowed. So now we the same process as in number a) just with a few adjustments.
x (1000
- 2.5
) = 997.5g
So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.
100g + 997.5g + 20g = 1117.5g or 1.1175kg
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great