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3 years ago

When a rigid body rotates about a fixed axis, all the points in the body have the same

Physics

1 answer:

timama [110]3 years ago

5 0

Hi there!

$\large\boxed{\text{Angular acceleration.}}$

For a rigid body rotating about a fixed axis, its angular characteristics ⇒ angular acceleration and velocity are constant throughout.

However, its LINEAR velocities/accelerations differ because of the following relationships:

v = ωr

a = αr

Thus, a point closer to the axis of rotation has a smaller linear velocity or acceleration compared to a point along the edge.

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A roller coaster is travelling 1 m/s at the top of the track. At the bottom of the track, 5 seconds later, it is travelling 36 m

lidiya [134]

Answer:

7m/s²

Explanation:

Given parameters:

Velocity at the top = 1m/s

Velocity at the bottom = 36m/s

Time = 5s

Unknown:

Average acceleration = ?

Solution;

Acceleration is the rate of change of velocity with time. It is expressed as;

A = $\frac{v -u}{t}$

v is the velocity at the top

u is the velocity at the bottom

t is the time taken

Now, insert the parameters and solve;

A = $\frac{36 - 1}{5}$ = 7m/s²

7 0

3 years ago

If you used 650 W of power lifting a 300 N weight in 2 seconds how high did you lift?

dimulka [17.4K]

Answer:

4.33m

Explanation:

Power = work done/ time

work done = power × time =650 × 2 = 1300J

work done = force × distance

distance = work done/force

distance = 1300/300 = 4.33m

7 0

3 years ago

An eartly method of measuring the speed of light makes use of a rotating slotted wheel.

Pavel [41]

Answer:

Explanation:

Distance traveled by light = 2 x 550 m = 1100 m

time taken by light to travel this distance = 1100 / 3 x 10⁸

= 366.67 x 10⁻⁸ s

angle between two consecutive slots = 2π / 500 rad

= .004π

angular velocity of wheel = angle moved / time taken

= .004π / 366.67 x 10⁻⁸

= 1091π radian / s

b ) linear speed of a point on the edge of the wheel

= ω R , r is radius of wheel , ω is angular velocity.

= 1091π x 5 x 10⁻²

= 1712.8 m /s

4 0

4 years ago

What is the work done by a car's braking system when it slows the 1500-kg car from an initial speed of 96 km/h down to 56 km/h i

kondor19780726 [428]

Answer:

-352.275KJ

Explanation:

We are given that

Mass of car=1500kg

Initial speed of car =u=96 km/h= $96\times \frac{5}{18}=26.67m/s$

1km/h= $\frac{5}{18}m/s$

Final speed of car=v=56km/h= $56\times \frac{5}{18}=15.56m/s$

Distance traveled by car=s=55m

We have to find the work done by the car's braking system.

Using third equation of motion

$v^2-u^2=2as$

$(15.56)^2-(26.67)^2=2a(55)$

$-469.18=110a$

$a=\frac{-469.18}{110}=-4.27m/s^2$

Where negative sign indicates that velocity of car decreases.

Work done by a car's barking system= $w=F\times s=ma\times s$

Work done by a car's barking system= $1500\times -4.27\times 55=352275J$

Work done by a car's barking system= $-\frac{352275}{1000}=-352.275KJ$

1KJ=1000J

Where negative sign indicates that work done in opposite direction of motion.

7 0

4 years ago

Read 2 more answers

Please help!!!!!!!!!

aleksandr82 [10.1K]

1. This question asks about velocity, so A and B are not correct. The car's velocity after 15 s with acceleration 2.00 m/s² would be

(2.00 m/s²) (15 s) = 30 m/s

[D]

2. Because Ima is slowing down to a stop, the acceleration is negative. Let x be the displacement of her vehicle during this motion. Then

0² - (30.0 m/s)² = 2 (-8.00 m/s²) x

==> x = (30.0 m/s)²/(2 (8.00 m/s²)) = 56.25 m ≈ 65.3 m

[A]

3. Since acceleration is constant, the average velocity is exactly the average of the initial and final velocities:

(21.0 m/s + 0 m/s)/2 = 10.5 m/s

The average (and thus instantaneous) acceleration during this time is equal to the change in velocity divided by the change in time:

(0 m/s - 21.0 m/s)/(6.00 s) = -3.50 m/s²

If x is the distance traveled as the car comes to a stop, then

0² - (21.0 m/s)² = 2 (-3.50 m/s²) x

==> x = (21.0 m/s)² / (2 (3.50 m/s²)) = 63.0 m

[A]

4.a. Assuming the sprinter's acceleration is constant, the average acceleration would be a such that

(11.5 m/s)² - 0² = 2 a (15.0 m)

==> a = (11.5 m/s)² / (2 (15.0 m)) ≈ 4.41 m/s²

4.b. By definition of average acceleration,

4.41 m/s² = (11.5 m/s - 0 m/s)/t

==> t = (11.5 m/s)/(4.41 m/s²) ≈ 2.61 s

5. At maximum height, any thrown object has zero velocity, so if it was thrown with an initial speed v, at its highest point we have

0² - v ² = 2 (-g) (91.5 m)

==> v = √(2g (91.5 m)) ≈ 42.3 m/s

(where I use g = 9.80 m/s²)

6.a. The brick's velocity after 7.0 s is

-g (7.0 s) = -68.6 m/s ≈ -69 m/s

6.b. The brick is presumably dropped from rest, so it is displaced by x such that

(-68.6 m/s)² - 0² = 2 (-g) x

==> x = -240.1 m ≈ -240 m

which is to say it falls a distance of 240 m. (The displacement is negative because we take its initial position to be the origin, and I took the downward direction to be negative.)

3 0

3 years ago