Answer:
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms
2 ).
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
23 =20 3 N
While the uertical component of the force acting in upward direction=Fsin30=40× 21
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN = 528.64
=5.73m/s 2
Ciara is winging....etc
The answer is : 0.60 N, toward the center of the circle
A satellite....etc
The Answer is : 7400 m/s
What is the .....etc
The Answer is : 2.60 m/s
Answer:
All i kno is that that kid ain't gonna be ok
Explanation:
if u tell me how to do it ill do it
Answer:
The approximate change in entropy is -14.72 J/K.
Explanation:
Given that,
Temperature = 22°C
Internal energy 
Final temperature = 16°C
We need to calculate the approximate change in entropy
Using formula of the entropy
Where, 
= internal energy
T = average temperature
Put the value in to the formula
Hence, The approximate change in entropy is -14.72 J/K.
