solution:
long to travel 19.6m at 28m/s
t
=19.6/28
=time taken to drop from rest
s=ut+(1/2)at^2
u=0
a=~10
so s=(1/2)*10*(19.6/28)^2 =2.45m
Answer: 2.5N
Explanation:
Given the following :
Mass of block (m) = 2kg
Coefficient of static friction (μs) = 0.4
Horizontal force applied to the block = 2.5N
The frictional force (Ff) between the block and the floor is :
First calculate the maximum static frictional force:
Frictional force = Coefficient of static friction(μs) × normal reaction(R)
Normal reaction(R) = mass × acceleration due to gravity (10m/s^2)
R = 2 × 10 = 20
Fmax = μs × R
Fmax = 0.4 x 20 = 8N
Here, since the applied force (2.5N) is less than maximum frictional force(8N).
The force of friction between the block and the floor will be equal to the applied force of 2.5N due its ability to adjust itself in other to ensure equilibrium.
Answer:
The current is 0.67 A.
Explanation:
Density, J = 1 A/m^2
Area, A = 1 m^2
Let the radius is r. And outer is R.
Use the formula of current density

Answer and work is shown in the image attached.
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Answer:
T = 2.82 seconds.
The frequency 
Amplitude A = 25.5 cm
The maximum speed of the glider is 
Explanation:
Given that:
the time taken for 11 oscillations is 31 seconds ;
SO, the time taken for one oscillation is :

T = 2.82 seconds.
The formula for calculating frequency can be expressed as :



The amplitude is determined by using the formula:

The limits that the spring makes the oscillations are from 10 cm to 61 cm.
The distance of the glider is, d = (61 - 10 )cm = 51 cm
Replacing 51 for d in the above equation

A = 25.5 cm
The maximum speed of the glider is:

where ;





