The motorist travels (a) 58 km/h and (b) ~16.1 m/sec
Answer:
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Explanation:
Answer:
7.08 m/s²
Explanation:
Given:
v₀ = 20.0 m/s
v = 105 m/s
t = 12.0 s
Find: a
v = at + v₀
105 m/s = a (12.0 s) + 20.0 m/s
a = 7.08 m/s²
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
Answer:
The third particle should be at 0.0743 m from the origin on the negative x-axis.
Explanation:
Let's assume that the third charge is on the negative x-axis. So we have:
We know that the electric field is:
Where:
- k is the Coulomb constant
- q is the charge
- r is the distance from the charge to the point
So, we have:
Let's solve it for r(3).
Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.
I hope it helps you!
