Given what we know, we can confirm that amongst the wave types mentioned both X-rays and radio waves are able to transmit energy through a vacuum.
<h3>How can these waves propagate through a vacuum?</h3>
This is due to the fact that unlike sound, water, and seismic waves that transmit energy through air, water, or the earth, the waves mentioned above, those being radio and x-rays, do not need a medium through which to propagate. This allows them to exist and transmit energy even in a vacuum such as outer space, which is corroborated by their existence in space.
Therefore, we can confirm that amongst the wave types mentioned both X-rays and radio waves are able to transmit energy through a vacuum.
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Answer:
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
Explanation:
Given that
g₁ = 9.96 m/s²
g₂ = 9.72 m/s²
The actual value of g = 9.8 m/s²
a)
The difference Δ g = 9.96 -9.72 =0.24 m/s²
![The\ percentage\ difference=\dfrac{0.24}{9.72}\times 100=2.46\ percentage\\](https://tex.z-dn.net/?f=The%5C%20percentage%5C%20difference%3D%5Cdfrac%7B0.24%7D%7B9.72%7D%5Ctimes%20100%3D2.46%5C%20percentage%5C%5C)
b)
For first one :
![Error\ in\ the\ percentage =\dfrac{9.96}{9.81}\times 100 =101.52\ perncetage](https://tex.z-dn.net/?f=Error%5C%20in%5C%20the%5C%20percentage%20%3D%5Cdfrac%7B9.96%7D%7B9.81%7D%5Ctimes%20100%20%3D101.52%5C%20perncetage)
For second :
![Error\ in\ the\ percentage =\dfrac{9.72}{9.81}\times 100 =99.08\ perncetage](https://tex.z-dn.net/?f=Error%5C%20in%5C%20the%5C%20percentage%20%3D%5Cdfrac%7B9.72%7D%7B9.81%7D%5Ctimes%20100%20%3D99.08%5C%20perncetage)
c)
The mean g(mean )
![g(mean )=\dfrac{9.96+9.72}{2}\ m/s^2\\g(mean)=9.84\ m/s^2](https://tex.z-dn.net/?f=g%28mean%20%29%3D%5Cdfrac%7B9.96%2B9.72%7D%7B2%7D%5C%20m%2Fs%5E2%5C%5Cg%28mean%29%3D9.84%5C%20m%2Fs%5E2)
![The\ percentage=\dfrac{9.84}{9.8}\times 100=100.40\ percentage](https://tex.z-dn.net/?f=The%5C%20percentage%3D%5Cdfrac%7B9.84%7D%7B9.8%7D%5Ctimes%20100%3D100.40%5C%20percentage)
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
Power Rating =
W
Time for which the machine is operating = 2 hours
Time in seconds = 2 × 60 × 60
Time = 7200 seconds
Now, we need to use the formula for work done.
Work done = Power × time
Work done = ![\( 1.1 \times 10^3 \times 7200\)](https://tex.z-dn.net/?f=%5C%28%201.1%20%5Ctimes%2010%5E3%20%5Ctimes%207200%5C%29)
Work done =
Joules
Hence, work done by the machine in 2 hours is
J.
Answer:
D. The net force on an object applied over a distance
Explanation:
Work can be defined as the product of force multiplied by a distance
Where
F = force [Newtons]
d = distance = [meters]
Therefore:
Work = F * d
Work in units of Joules [J]
E=mcθ
where E is the energy added,m is the mass,c is the specfic heat capacity and θ is the change in temprature.
making m subject u get E/c<span>θ=m
</span><span>θ=(30-20)=10</span><span>
plugging the values we get :
</span>
![\frac{4750}{1900*10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4750%7D%7B1900%2A10%7D%20)
<span>
solving it we get the answer that is 0.25kg or 250 grams
</span>