The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.
The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.
v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s
Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.
The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.
Combining this together we get:
(1) vx=40m/s and vy=10m/s
Answer:
(A) Q = 321.1C (B) I = 42.8A
Explanation:
(a)Given I = 55A−(0.65A/s2)t²
I = dQ/dt
dQ = I×dt
To get an expression for Q we integrate with respect to t.
So Q = ∫I×dt =∫[55−(0.65)t²]dt
Q = [55t – 0.65/3×t³]
Q between t=0 and t= 7.5s
Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]
Q = 321.1C
(b) For a constant current I in the same time interval
I = Q/t = 321.1/7.5 = 42.8A.
Answer:
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The atoms and molecules in it are in constant motion. The kinetic energy of such a body is the measure of its temperature. Potential energy is classified depending on the applicable restoring force. Gravitational potential energy – potential energy of an object which is associated with gravitational force
in cgs system, plank's constant= h=6.626 x10⁻²⁶ erg s
Value of Plank's constant in SI system= 6.626 x10⁻³⁴ Js
now 1 Joule= 10⁷ ergs
so h= 6.626 x10⁻³⁴ Js (10⁷ ergs/1J)
h=6.626 x10⁻²⁷ erg s
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