Your answer is electricity, light and magnetism. They can be determined usinf elecromagnetic radioation.
<span>
Even the energy can't be detected by our eyes, there are a lot of measurement instruments that can measure infrared (IR), gamma rays, radio or X-rays or ultraviolet (UV)</span>
"<span>The current is the same at all points" is the one among the following choices given in the question that answers the question correctly. The correct option among all the options that are given in the question is the fifth option or the last option. I hope that this is the answer that has come to your desired help.</span>
Answer:
1020g
Explanation:
Volume of can=![1100cm^3=1100\times 10^{-6}m^3](https://tex.z-dn.net/?f=1100cm%5E3%3D1100%5Ctimes%2010%5E%7B-6%7Dm%5E3)
![1cm^3=10^{-6}m^3](https://tex.z-dn.net/?f=1cm%5E3%3D10%5E%7B-6%7Dm%5E3)
Mass of can=80g=![\frac{80}{1000}=0.08kg](https://tex.z-dn.net/?f=%5Cfrac%7B80%7D%7B1000%7D%3D0.08kg)
1Kg=1000g
Density of lead=![11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3](https://tex.z-dn.net/?f=11.4g%2Fcm%5E3%3D11.4%5Ctimes%2010%5E%7B3%7D%3D11400kg%2Fm%5E3)
By using ![1g/cm^3=10^3kg/m^3](https://tex.z-dn.net/?f=1g%2Fcm%5E3%3D10%5E3kg%2Fm%5E3)
We have to find the mass of lead which shot can it carry without sinking in water.
Before sinking the can and lead inside it they are floating in the water.
Buoyancy force =![F_b=Weight of can+weight of lead](https://tex.z-dn.net/?f=F_b%3DWeight%20of%20can%2Bweight%20of%20lead)
![\rho_wV_cg=m_cg+m_lg](https://tex.z-dn.net/?f=%5Crho_wV_cg%3Dm_cg%2Bm_lg)
Where
Density of water
Mass of can
Mass of lead
Volume of can
Substitute the values then we get
![1000\times 1100\times 10^{-6}=0.08+m_l](https://tex.z-dn.net/?f=1000%5Ctimes%201100%5Ctimes%2010%5E%7B-6%7D%3D0.08%2Bm_l)
![1.1-0.08=m_l](https://tex.z-dn.net/?f=1.1-0.08%3Dm_l)
![m_l=1.02 kg=1.02\times 1000=1020g](https://tex.z-dn.net/?f=m_l%3D1.02%20kg%3D1.02%5Ctimes%201000%3D1020g)
![1 kg=1000g](https://tex.z-dn.net/?f=1%20kg%3D1000g)
Hence, 1020 grams of lead shot can it carry without sinking water.
Answer:![u=\frac{v}{2}\sqrt{5-4sin\phi }](https://tex.z-dn.net/?f=u%3D%5Cfrac%7Bv%7D%7B2%7D%5Csqrt%7B5-4sin%5Cphi%20%7D)
Explanation:
Given
Both cars mass is m
and solving problem in Vertical and horizontal direction
considering + y and +x to be positive and u be the final velocity of system
Conserving Momentum in Vertical direction
![m(2v)+m(-vsin\phi )=2m(ucos\theta )](https://tex.z-dn.net/?f=m%282v%29%2Bm%28-vsin%5Cphi%20%29%3D2m%28ucos%5Ctheta%20%29)
------1
Conserving momentum in x direction
-----2
squaring and adding 1 &2
![(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2](https://tex.z-dn.net/?f=%282u%29%5E2%3D%282v-vsin%5Cphi%20%29%5E2%2B%28vcos%5Cphi%20%29%5E2)
![4u^2=4v^2+v^2-4v^2sin\phi](https://tex.z-dn.net/?f=4u%5E2%3D4v%5E2%2Bv%5E2-4v%5E2sin%5Cphi%20)
![4u^2=5v^2-4v^2sin\phi](https://tex.z-dn.net/?f=4u%5E2%3D5v%5E2-4v%5E2sin%5Cphi%20)
![u=\frac{v}{2}\sqrt{5-4sin\phi }](https://tex.z-dn.net/?f=u%3D%5Cfrac%7Bv%7D%7B2%7D%5Csqrt%7B5-4sin%5Cphi%20%7D)
Answer:
d
Explanation:
the heat can move through the metal and onto the clothing