Response:
0.60 m/s
Details:
The average speed between times t = a and t = b can be expressed as:
v_avg = (x(b) − x(a)) / (b − a)
Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:
v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)
v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0
v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0
v_avg = [0.96 − (-0.84)] / 3.0
v_avg = 0.60
The average speed calculated is 0.60 m/s.
Answer:
The flux across the cube's surface is 
.
Solution:
According to the details provided:
Cube edge length, a = 8.0 cm = 
.
Volume charge density, 
.
Now,
To find the electric flux:
where
= electric flux
= permittivity of vacuum.
The volume charge density for this scenario is described by:
Cube volume, 
.
Thus,
.
The total charge can be derived from equation (2):
.
.
Now, insert the value of 'q' into equation (1):
.
Answer:
All observers are accurate.
Explanation:
This situation reflects a matter of reference frames regarding the book's motion as perceived by different observers.
From their distinct frames of reference, each observer's perspective is valid.
Observer A is in an inertial reference frame.
Observers capable of explaining the book's behavior and its relationship to the car through the interplay of forces and changes in velocity are classified as being in inertial reference frames.
Observer A's observations illustrate this, for she pointed out the relative motion between the book and the car, indicating her position in an inertial reference frame.
Likewise, observers in these inertial reference frames can elucidate object velocity changes based on the forces affecting them from other objects.
This is exemplified by observer B, who notes the car's force impacting the book's velocity.
Observer C occupies a non-inertial reference frame, as Newton's laws of motion do not apply. This scenario arises within non-inertial frames.
Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
The velocity at the stagnation point is zero.
The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Thus, the answer is 2.2 psi
Answer:
Explanation:
The beacon is rotating at an angular speed of
so we have
We know that
At this point we have
So we can conclude with
