Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
Answer:
D
Explanation:
you can catch it, rolling against a surface can slow it and when you throw it in the air, gravity will slow it down until it drops
Answer / Explanation
It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.
So for proper understanding, the two parts of the question will be highlighted.
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.
a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?
b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.
Solve for Vmax
Step Process
a) Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]
Where V₀ = 6m/s,
We have,
h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]
= 0.411 m
Where V₀ = 9.5m/s
We have,
h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]
= 1152 m
b) From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.
Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.
Therefore Vₙ = V₀max = 9.8 m/s
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
