Answer:
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
Explanation:
Hi there!
The equation of velocity of the falling egg is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time
Let´s calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².
v = v0 + a · t
11.1 m/s = 0 m/s + a · 1.13 s (since the egg is dropped, the initial velocity is zero). Solving for "a":
11.1 m/s / 1.13 s = a
a = 9.82 m/s²
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
Now, using the same equation, let´s find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:
v = v0 + a · t
0 = 11.1 m/s + a · 0.140 s
-11.1 m/s / 0.140 s = a
a = -79.3 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
The instruments would not produce sound waves therefore make no noise
Answer:
Mechanical advantage = 3
Explanation:
You exert a 100-N force on a pulley system to lift 300-N.
The mechanical advantage of the system is given by the ratio of output force to the input force.
Here, output force = 300 N and input force = 100 N
Mechanical advantage,
Mechanical advantage is 3 it means that there are 3 sections of rope support. Hence, this is the required solution.
Assuming air as ideal gas and amount of air in no of moles is known then by gas law,
PV= nRT
Pressure is constant
P* (change in volume) = nR* (change in temperature)
Answer:
(a) its moment of inertia about its center is 0.002095 kgm²
(b) Applied torque is 0.071813 Nm
Explanation:
Given;
Radius of the grinding wheel, R = 8.5cm
Mass of the grinding wheel, m = 0.580kg
Part (a) its moment of inertia about its center
I = ¹/₂MR²
I = ¹/₂(0.58)(0.085)²
I = 0.002095 kgm²
Part (b)
Given;
initial angular velocity, ωi = 1500rpm = 157.1 rad/s
final angular velocity, ωf = 1500rpm = 157.1 rad/s
Initial torque, τi = I x αi
αi = ωi / t
αi = 157.1 / 5 = 31.42 rad/s²
τi = 0.002095 x 31.42
τi = 0.06583 Nm
Final torque, τf = I x αf
αf = ωf / t
αf = 157.1 / 55 = 2.856 rad/s²
τf = 0.002095 x 2.856
τf = 0.005983 Nm
Applied torque = τi + τf
= 0.06583 Nm + 0.005983 Nm
= 0.071813 Nm