Answer:
Area=1.5(1.5)=2.25m^2
Force of gravity=10N
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}
⟼Pressure=
Area
Force
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}
⟼Pressure=
2.25
10
\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}
⟼Pressure=4.4Pa
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Answer:
Angular acceleration = 23.68 rad / s²
Explanation:
Given that,
acceleration = 9m/s²
Therefore acceleration of string is 9m/s²
since string is constant in length
cylinder of radius 38.0 cm = 0.38m
Angular acceleration = a / r
Angular acceleration = 9 / 0.38
= 23.68 rad / s²
Angular acceleration = 23.68 rad / 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
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