Answer:
The fraction of its volume inside liquid is increased .
Explanation:
According to principle pf floatation , an object floats on the surface of water
when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .
In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .
Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
Because an object in rest stays in rest until an unequal force pushes it so gravity is pushing on the egg making it drop
The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
Learn more about kinetic friction here-
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Answer:
400m
Explanation:
Brainliest? :))
Let your initial displacement from your home to the store be
Dd
>
1 and your displacement from the store to your friend’s house
be Dd
>
2.
Given: Dd
>
1 = 200 m [N]; Dd
>
2 = 600 m [S]
Required: Dd
>
T
Analysis: Dd
>
T 5 Dd
>
1 1 Dd
>
2
Solution: Figure 6 shows the given vectors, with the tip of Dd
>
1
joined to the tail of Dd
>
2. The resultant vector Dd
>
T is drawn in red,
from the tail of Dd
>
1 to the tip of Dd
>
2. The direction of Dd
>
T is [S].
Dd
>
T measures 4 cm in length in Figure 6, so using the scale of
1 cm : 100 m, the actual magnitude of Dd
>
T is 400 m.
Statement: Relative to your starting point at your home, your
total displacement is 400 m [S].
