Answer:
A×B=C×D
500×0.5=250×X
250=250×X
X=250/250=1
X=1 m
Explanation:
note: if the force plus two, the distance will be half.
In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?
<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.
Answer:
Refer to the attachment for solution (1).
<h3><u>Calculating time taken by it to stop (t) :</u></h3>
By using the second equation of motion,
→ v = u + at
- v denotes final velocity
- u denotes initial velocity
- t denotes time
- a denotes acceleration
→ 0 = 5 + (-5/6)t
→ 0 = 5 - (5/6)t
→ 0 + (5/6)t = 5
→ (5/6)t = 5
→ t = 5 ÷ (5/6)
→ t = 5 × (6/5)
→ t = 6 seconds
→ Time taken to stop = 6 seconds
Answer:
-1m/s
Explanation:
We can calculate the speed of block A after collision
According to collision theory:
MaVa+MbVb = MaVa+MbVb (after collision)
Substitute the given values
5(3)+10(0) = 5Va+10(2)
15+0 = 5Va + 20
5Va = 15-20
5Va = -5
Va = -5/5
Va = -1m/s
Hence the velocity of ball A after collision is -1m/s
Note that the velocity of block B is zero before collision since it is stationary
Let t = Theta and p = Phi
Tan t = y/x Then x =y/Tant.
Tant = y/(x-d) x-d = y/Tanp
y/Tant - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp = d*Tanr
y(1 - Tanr/Tanp = d*Tant
y = d*Tant/(1-Tant/Tanp)
