Complete Question:
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to
Answer:
t= 16.7 sec.
Explanation:
As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:
γ = (ωf -ω₀) / t
If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:
γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².
When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:
γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec
Answer:
A) and B) are correct.
Explanation:
If the object is at rest, it means that no net force is exerted on it.
As the object experiences a downward gravitational force from Earth, in order to be at rest, it must experience an upward force with the same magnitude as the gravitational force on the object.
This force is supplied by the normal force, which can adopt any value in order to meet the condition imposed by Newton´s 2nd Law, and is always perpendicular to the surface on which the object is placed (in this case, the ground).
At a molecular level, this normal force is supplied by the bonded molecules of the ground that behave like small springs being compressed by the molecules of the object, exerting an upward restoring force upward on them.
So, the statements A) and B) are true.
Answer:
Explanation:
From the question we are told that:
Time 
Generally the Period is given as
Therefore difference in frequency dF
Answer:
169.98 m
Explanation:
= Thermal coefficient of marble = 
= Initial length = 170 m
= Change in temperature = (-10-35)°C
Change in length is given by
The height on the given day will be 
Answer:
R = 6.3456 10⁴ mile
Explanation:
For this exercise we will use Newton's second law where force is gravitational force
F = m a
The satellite is in a circular orbit therefore the acceleration is centripetal
a = v² / r
Where the distance is taken from the center of the Earth
G m M / r² = m v² / r
G M / r = v²
The speed module is constant, let's use the uniform motion relationships, with the length of the circle is
d = 2π r
v = d / t
The time for a full turn is called period (T)
Let's replace
G M / r = (2π r / T)²
r³ = G M T²²2 / 4π²
r = ∛ (G M T² / 4π²)
We have the magnitudes in several types of units
T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s
Re = 6.37 10⁶ m
Let's calculate
r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)
r = ∛ (1.027487 10²⁴)
r = 1.0847 10⁸ m
This is the distance from the center of the Earth, the distance you want the surface is
R = r - Re
R = 108.47 10⁶ - 6.37 10⁶
R = 102.1 10⁶ m
Let's reduce to miles
R = 102.1 10⁶ m (1 mile / 1609 m)
R = 6.3456 10⁴ mile
