Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:






we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:



Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:







t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:



So once we got the two distances we can now find the difference between them:

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Answer:E
Explanation:
It is given that Energy of gamma ray is E=1.2 Mev
Shielding effect can be measured by measuring the fraction of gamma rays blocked by shield. If certain thickness will able to block half the radiation then to block 75% radiation we need to add same amount of thickness in order to block the remaining radiation.
i.e.
fraction is blocked by 10 cm thickness
then remaining radiation is 
another 10 cm thickness will block the remaining half radiation i.e.
so total 75 % radiation will be blocked
Answer:
Change in temperature ∆(tita) is 266.097°C
Explanation:
Ok kinectic energy = 1/2MV²
5.4 grams =( 5.4/1000) kilogram
Kinectic energy =( 1/2 )*(5.4/1000)*261²
Kinectic energy = 183.9267 joules
If kinetic energy = thermal energy
183.9267 joules = mc∆(tita)
Where ∆(tita) = change in temperature
And c = 128 J/kg
∆(tita) = 183.9267/((5.4/1000)*128)
∆(tita) = 266.097
∆(tita) = 266.097°C
Torque = r x F
|F| = mg = 60 * 10 N = 600 N ( assuming g ~ 10m/s^2)
distance of fulcrum = torque / Force = 90/600 m = .15 m.
Answer: Well you didn't give any answers so my guess would be it depends how much force you put into it and where you throw it.
Explanation: