Answer:
3.43 m
Explanation:
f = 100 mm
u = - 103 mm
Let v be the distance between the screen and the lens of the projector.
Use lens equation
1 / f = 1 / v - 1 / u
1 / 100 = 1 / v + 1 / 103
1 / v = 1 / 100 - 1 / 103
1 / v = (103 - 100) / (100 x 103)
1 / v = 3 / 10300
v = 3433.33 mm = 3.43 m
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Answer:
t = 4.17 [s]
Explanation:
We know that work is defined as the product of force by distance.
W = F*d
where:
F = force [N] (units of Newtons)
d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]
In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.
w = m*g
where:
m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]
g = gravity acceleration = 9.81 [m/s²]
w = 14.7*9.81
w = 144.2 [N]
Therefore the work can be calculated.
W = w*d
W = 144.2*63.4
W = 9142.72 [J] (units of Joules)
Power is now defined in physics as the relationship of work at a given time
P = W/t
where:
P = power = 2190 [W]
t = time [s]
Now clearing t, we have.
t = W/P
t = 9142.72/2190
t = 4.17 [s]
Answer:
F = 0.535 N
Explanation:
Let's use the concepts of energy, at the highest and lowest point of the trajectory
Higher
Em₀ = U = mg y
Lower
= K = ½ m v²
Emo =
mg y = ½ m v2
v = √ 2gy
y = L - L cos θ
v = √ (2g L (1-cos θ))
Now let's use Newton's second law n at the lowest point where the acceleration is centripetal
F = ma
a = v² / r
In turning radius is the cable length r = L
F = m 2g (1-cos θ)
Let's calculate
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
Answer:
F = - 59.375 N
Explanation:
GIVEN DATA:
Initial velocity = 11 m/s
final velocity = 1.5 m/s
let force be F
work done = mass* F = 4*F
we know that
Change in kinetic energy = work done
kinetic energy = 
kinetic energy = 
= -237.5 kg m/s2
-237.5 = 4*F
F = - 59.375 N
