Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules
As a horse finishes its trot around the corral, it slows from 4m/s to a stop in 3
seconds. Calculate the acceleration of the horse.
Can anybody help ?
Answer:a) 20.944rad/s (b) 12.57m/s (c) 800rev/min²
Explanation:
Answer:
Let, R be the resistance of the heater wire. Since, two heater wires are of equal length their resistance is also same.
Hence, for series combination of resistances,
R
s
=2R
And for parallel combination of resistances,
R
p
=
2
R
Now, heat produced when they first connected in series is
H
s
=
R
s
V
2
where, V is voltage supplied to the heater.
H
s
=
2R
V
2
.....................................(1)
heat produced when they first connected in series is
H
p
=
R
p
V
2
H
p
=
2
R
V
2
H
p
=
R
2V
2
............................................(2)
From (1) and (2), we get
H
p
H
s
=
R
2V
2
2R
V
2
⇒
H
p
H
s
=
4
1
