The driver speeds up with acceleration <em>a</em> so that
35 m/s = 15 m/s + <em>a</em> (10.0 s)
Solve for <em>a</em> :
20 m/s = <em>a</em> (10.0 s)
<em>a</em> = (20 m/s) / (10.0 s)
<em>a</em> = 2 m/s²
Answer:
5000 kg/m^3
Explanation:
Here. we are asked to calculate the density of the rock specimen.
we proceed as follows;
mass of water displaced is calculated by finding the difference between the actual and apparent masses
This has a value of 0.45kg - 0.36kg = 0.09kg
The rock and water that is displaced have exactly the same volume and thus their densities is the same. This makes the ratio of their masses to be the same
Ratio of masses is
0.45 / 0.09 = 5.0
Here we can see that the mass of the rock is five times the mass of the water so it must be five times denser
Thus, since the density of water is 1000 kg/m^3 , the density of rock is 5000 kg/m^3
Answer:
Explanation:
mass, m = 50 g
moment of inertia, I = 9 x 106-5 kg m^2
radius, r = 0.7 cm
(a) As it moving downwards
Let T be the tension in the string
T = m (g + a) .... (1)
where, a be the acceleration
τ = I α = T r
α = a / r
So, I x a / r = T x r
a = T r^2 / I
Substitute in equation (1) we get
a = m (g + a) r^2 / I
a = mgr^2 / (I - mr^2)
a = 0.050 x 9.8 x 0.007 x 0.007 / (9 x 10^-5 - 0.050 x 0.007 x 0.007)
a = 2.401 x 10^-5 / (87.55 x 10^-6)
a = 0.274 m/s^2
τ = I x α = I x a / r
τ = 9 x 10^-5 x 0.274 / 0.007
τ = 3.52 x 10^-3 Nm
(b) α = a / r
α = 0.274 / 0.007 = 39.14 rad/s^2
This drag force is always opposite to the object's motion, and unlike friction between solid surfaces, the drag force increases as the object moves faster.
Answer:
Pressure = Density x Gravity acceleration x Height
Density of water = 1000 Kgm^-3
Gravity acceleration = 9.8 ms^-2
So it is 4.50.
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