A body of volume 2000 cm3 has a mass 4 kg. Find the density of the body. Will the body sink or float in water, if the density of
1 answer:
Answer:
, it will sink
Explanation:
The density of an object is given by
where
m is the mass of the object
V is its volume
For the body in the problem, we have
m = 4 kg = 4000 g
Therefore, its density is
And the object will sink in water, because its density is larger than that of water, which is . (an object sinks when its density is larger than that of water, otherwise it floats).
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Answer:
The length at the final temperature is 11.7 cm.
Explanation:
We need to use the thermal expansion equation:
Where:
- L(0) is the initial length
- ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
- ΔL is the final length minus the initial length (L(f)-L(0))
- α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)
So, we have:
Therefore, the length at the final temperature is 11.7 cm.
I hope it helps you!
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² = / r
ω = √( / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849