A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
1 answer:
Answer:
Magnitude of the force is 2601.9 N
Explanation:
m = 450 kg
coefficient of static friction μs = 0.73
coefficient of kinetic friction is μk = 0.59
The force required to start crate moving is .
but once crate starts moving the force of friction is reduced .
Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie .
Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.
Magnitude of the force
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Answer:
Minimum number of photons required is 1.35 x 10⁵
Explanation:
Given:
Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m
Energy of one photon is given by the relation :
....(1)
Here h is Planck's constant and c is speed of light.
Let N be the minimum number of photons needed for triggering receptor.
Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J
According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,
E₁ = N x E
Put equation (1) in the above equation.
Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.
N = 1.35 x 10⁵
Explanation:
At point B, the velocity speed of the train is as follows.
=
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y =
Now, second derivative of the train is calculated as follows.
Radius of curvature of the train is as follows.
=
= 3808.96 m
Now, we will calculate the normal component of the train as follows.
=
= 0.1822
The magnitude of acceleration of train is calculated as follows.
a =
=
=
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is .