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dedylja [7]
3 years ago
9

uppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s

after 36.0 seconds have elapsed. How long after the initial time has the wheel undergone half of the angular displacement that it will have gone through during the entire 36.0-second interval
Physics
1 answer:
Paul [167]3 years ago
8 0

Answer:

t = 22.2 s

Explanation:

angular distance covered in the 36.0 s is

θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians

720/2 = 360 radians

α = Δω/t = (30 - 10)/36 = 5/9 rad/s²

θ = ω₀t + ½αt²

360 = 10.0t + ½(5/9)t²

   0 = (5/18)t² + 10.0t - 360

   0 = t² + 36t - 1296

t = (-36 ±√(36² - 4(1)(-1296))) / 2

t = (-36 ±√(6480)) / 2

t = -18 ±√1620

we ignore the negative time result as it occurs before we care.

t = -18 + √1620 = 22.249223... s

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5 0
3 years ago
An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car
Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)=3.2 m/s^2

Speed of car after 4.5 s

using equation of motion

v=u+at

v=4.92+3.2\times 4.5=4.92+14.4

v=19.32 m/s

Displacement of the car after 4.5 s

v^2-u^2=2as

19.32^2-4.92^2=2\times 3.2\times s

349.05=2\times 3.2\times s

s=54.54 m

4 0
2 years ago
Newtons laws of motion
Viktor [21]

Answer:

Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.

Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.

Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.

7 0
2 years ago
14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.
sineoko [7]

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

3 0
3 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
2 years ago
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