The braking force is -400 N
Explanation:
We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

where in this problem, we have:
F is the force applied by the brakes
is the time interval
m = 13,000 kg is the mass of the ferry
u = 2.0 m/s is the initial velocity
v = 0 is the final velocity
And solving for F, we find the force applied by the brakes:

where the negative sign indicates that the direction is backward.
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Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:

where
Answer:
Average density of Sun is 1.3927
.
Given:
Radius of Sun = 7.001 ×
km = 7.001 ×
cm
Mass of Sun = 2 ×
kg = 2 ×
g
To find:
Average density of Sun = ?
Formula used:
Density of Sun = 
Solution:
Density of Sun is given by,
Density of Sun = 
Volume of Sun = 
Volume of Sun = ![\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%203.14%20%5Ctimes%20%5B7.001%20%5Ctimes%2010%5E%7B10%7D%5D%5E%7B3%7D)
Volume of Sun = 1.436 ×

Density of Sun = 
Density of Sun = 1.3927 
Thus, Average density of Sun is 1.3927
.
Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision
hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)
Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %