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serious [3.7K]
3 years ago
9

A wheel rotates with a constant angular acceleration of 6.0 rad/s2. During a certain time interval its angular displacement is 8

.0 rad. At the end of the interval its angular velocity is 14.0 rad/s. What is its angular velocity at the beginning of the interval
Physics
1 answer:
kari74 [83]3 years ago
3 0

The angular velocity at the beginning of the interval of the wheel rotating  at a constant angular acceleration is determined as 10 rad/s.

<h3>Initial angular velocity of the wheel</h3>

The initial angular velocity of the wheel is determined by applying the kinematic equation as shown below;

ωf² = ωi² + 2αθ

where;

  • ωf is the final angular velocity
  • ωi is the initial angular velocity
  • α is angular acceleration
  • θ is angular displacement

Substitute the given parameters and solve for the initial angular velocity.

14² = ωi² + 2(6)(8)

196 = ωi² + 96

ωi² = 196 - 96

ωi² = 100

ωi = √100

ωi = 10 rad/s

Thus, the initial angular velocity of the wheel is 10 rad/s.

Learn more about angular velocity here: brainly.com/question/6860269

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If the speed of light in a substance is 2.26 x 10^8 m/s, what is the index of refraction of that substance?
vladimir1956 [14]

Answer:

1.33

Explanation:

speed of light in vacuum, c = 3 x 10^8 m/s

speed of light in medium, v = 2.26 x 10^8 m/s

The refractive index of the medium is given by

μ = speed of light in vacuum / speed of light in medium

μ = (3 x 10^8) / (2.26 x 10^8)

μ = 1.33

4 0
3 years ago
A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f
meriva

Answer:

v=\sqrt{\dfrac{10g(R-r)}{7}}

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

I=\dfrac{2}{5}mr^2

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for  rolling without slipping       v= ωr

Form energy conservation

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

 v= ωr

I=\dfrac{2}{5}mr^2

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2

mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2

2mg(R-r)=mv^2+\dfrac{2}{5}mv^2

2g(R-r)=\dfrac{7}{5}v^2

v=\sqrt{\dfrac{10g(R-r)}{7}}

3 0
3 years ago
A bullet is fired from a gun at 45° angle to the horizontal with a velocity of 500 m/s. Find the
rjkz [21]

answer :

D. 6370.92 m

Explanation:

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7 0
3 years ago
What U.S law related to drug use was passed in 1906
wolverine [178]

Answer:

Pure Food and Drug Act of 1906

Explanation:

The Pure Food and Drug Act of 1906 prohibited the sale of misbranded or adulterated food and drugs in interstate commerce and laid a foundation for the nation's first consumer protection agency, the Food and Drug Administration 

8 0
4 years ago
Read 2 more answers
why can we use P=VI in the question electric bulb base generator is rated 220 volts and 100W when it is operated on 110v what wi
MrRissso [65]

Answer:

Correct answer:  P₂ = 25 W

Explanation:

Given: voltage V₁ = 220 V, power P₁ = 100 W, V₂ = 110 V, P₂ = ?

The formula for calculating power is:

P = V · I

We will include in the story and ohm's law:

I = V/R

We will replace the current in the expression for power

P = V · V/R = V²/R  ⇒ R = V²/P

We will first calculate the electrical resistance of the bulb which is a constant in the electrical circuit

R = V₁²/P₁ = 220²/ 100 = 48,400 / 100 = 484 Ω

Power consumption of bulb connected to 110 V is:

P₂ = V₂²/R = 110²/484 = 12,100/484 = 25 W

P₂ = 25 W

God is with you!!!

8 0
3 years ago
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