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Cloud [144]
3 years ago
9

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period At 1.90 s, the tire is stopped and spun in the o

pposite (clockwise) direction also at 2.70 rad/s. Calculate the change in the tire's angular velocity Aw and the tire's average angular acceleration aav. (Indicate the direction with the signs of your answers.)
(a) the change in the tire's angular velocity Aco (in rad/s) rad/s
(b) the tire's average angular acceleration aav (in rad/s2) rad/s
Physics
1 answer:
Naddik [55]3 years ago
7 0

Answer:

a) -5.40 rad/s

b) -2.842 rad/s²

Explanation:

The direction is important in dealing with such questions. Clockwise is considered negative and counterclockwise is considered positive

a) Δω = final angular velocity - initial angular velocity

          = -2.70 rad/s - 2.70 rad/s

          = -5.40 rad/s

b) ∝ = Δω/Δt = (-5.40 rad/s)/1.90s = -2.842 rad/s²

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Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str
Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, v_y, of the package just before landing is given by the relation;

v_y² = u² + 2·g·h

u = 0 m/s

∴ v_y² = 0 + 2×9.81×1500 = 29430 m²/s²

v_y = √29430  = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

tan \theta = \dfrac{v_y}{v_x}  = \dfrac{171.55}{39.0 } = 4.4

∴ θ = tan⁻¹(4.4) = 77.19°.

6 0
3 years ago
3. A bicycle has a momentum of 25.00 kg* m/s and a velocity of 2.5 m/s . What is the bicycle's
Anna35 [415]

Answer:

10 kg

Explanation:

The question is most likely asking for the mass of the bicycle.

Momentum is the product of an object's mass and velocity. Mathematically:

p = m * v

Where p = momentum

m = mass

v = velocity

Hence, mass is:

m = p / v

From the question:

p = 25 kgm/s

v = 2.5 m/s

Mass is:

m = 25 / 2.5 = 10 kg

The mass of the bicycle is 10 kg.

In case the question requires the Kinetic energy of the bicycle, it can be gotten by using the formula

K. E = ½ * p * v

K. E. = ½ * 25 * 2.5 = 31.25 J

5 0
3 years ago
According to Boyle’s Law, when pressure is increased...
klasskru [66]

Boyle's law states that when you "shrink"(change its size) a container the pressure increases so the best answer would be D-volume is decreased.

4 0
3 years ago
A particular car has a weight of 9500 N. What is the mass of the car? 9500 kg 95000 kg 950 kg 2160 kg
Ivenika [448]

950kg. It is W=mg. therefore, mass= W/g which is W/10

=9500/10

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8 0
3 years ago
I don’t know the answer <br><br> Help me please
lubasha [3.4K]

Answer:

Explanation:

i wanna say its B but if im wrong sorry i tried

5 0
3 years ago
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