From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
= 2×9.81×0.43
= 8.4366
u = √8.4366
=2.905 m/s
Hence the initial velocity is 2.905 m/s
Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
= 2.905/9.81
= 0.296 seconds
The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.
<h3>Final speed of the nickel</h3>
Apply the principle of conservation of energy.
Q = mcΔθ
Q = (18)(0.444)(66 - 20)
Q = 367.63 J
Q = K.E = ¹/₂mv²
2K.E = mv²
v = √(2K.E/m)
where;
v = √(2 x 367.63)/(0.018))
v = 202.1 m/s
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Answer:
option (b)
Explanation:
Let the resistance of each resistor is R.
In series combination,
The effective resistance is Rs.
rs = r + R + R + .... + n times = NR
Let V be the source of potential difference.
Power in series
Ps = v^2 / Rs = V^2 / NR ..... (1)
In parallel combination
the effective resistance is Rp
1 / Rp = 1 / R + 1 / R + .... + N times
1 / Rp = N / R
Rp = R / N
Power is parallel
Rp = v^2 / Rp = N V^2 / R ..... (2)
Divide equation (1) by equation (2) we get
Ps / Pp = 1 / N^2
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s
2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration
3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.
So the maximum acceleration on the block is
4)As the box slides, it is now subjected to kinetic friction, which is
So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2