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GaryK [48]
4 years ago
8

Which layers of the earth are considered liquid a) crust b) inner corec) outer core d) lower mantle

Physics
1 answer:
Dima020 [189]4 years ago
3 0

the answer is d) outer core.

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A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.
REY [17]

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}} $

  $= \frac{g \sin \theta}{1 +\frac{2}{3}}$

  $= \frac{3}{5} g \sin \theta$

  = 0.6 x 9.81 x sin ( 29.7)

  $= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

$\Delta t = \sqrt{\frac{2 \times 2.9}{2.913}}$

Δt = 1.411 s

6 0
3 years ago
A car goes East from 2 m/s to 16m/s in 3.5s. What is the Car's acceleration?
BaLLatris [955]
A=v1+v2 / t
2+16/3.5=?
8 0
3 years ago
Describe the agronomic significance of the upper and lower plastic limits, and of the plasticity index?​
GarryVolchara [31]

Answer:It shows the size of the range of the moisture contents at which the soil remains plastic. In general, the plasticity index depends only on the amount of clay present. It indicates the fineness of the soil and its capacity to change shape without altering its volume.

4 0
3 years ago
A bulldozer and a Mini Cooper are involved in a head-on collision. Which one experiences a greater force
Pepsi [2]

Answer:

The mini Cooper will experience the greater force

Explanation:

Generally, a bulldozer has a greater mass compared to a Mini Cooper hence when both of these vehicles interact in an head on collision the Mini Cooper will experience a greater force because the bulldozer has a greater momentum

5 0
3 years ago
The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?
anastassius [24]
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
\rho =  \frac{AR}{L}
where
\rho is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
A=\pi r^2
where r is the radius of the wire. Substituting in the previous equation ,we find
\rho =  \frac{\pi r^2 R}{L}

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
\rho' =  \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4  \frac{\pi r^2 R}{L}   = 4 \rho
Therefore, the new resistivity must be 4 times the original one.
5 0
3 years ago
Read 2 more answers
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