4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]

Answers

Answer 1
Answer:

The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.

What is angular velocity?

The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.

ω = θ t

Where,

θ is the angle of rotation,

tis the time

ω is the angular velocity

The given data in the problem is;

u is the initialvelocity=0

α is the angularacceleration =  4.0 rad/s²

t is the time period=

n is the number of revolution = 10 rev

From Newton's second equation of motion in terms of angular velocity;

\rm \omega_f^2 - \omega_i^2 = 2as \n\n \rm \omega_f^2 - 0 = 2* 4 * 62.83 \n\n \rm \omega_f= 22.41 \ rad/sec

Hence the angular velocity will be 22.41 rad/s.

To learn more about angularvelocity refer to the link

brainly.com/question/1980605

Answer 2
Answer:

Answer:

w_f= 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

w_f^2-w_i^2=2aS

where w_f is the final angular velocity, w_i the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

w_f^2-(0)^2=2(4)(62.83)

Finally, solving for w_f:

w_f= 22.41rad/s


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An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?

The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7 T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.

Answers

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed  ,E=Electric filed

Power  P = I A

A=Area=4πr²  ,I=Intensity

I=(CB^2)/(2\mu_0)

I=(CE^2)/(2\mu_0 C^2)

E=\sqrt{{2I\mu_0 C}}

E=\sqrt{{2* 1150* 4\pi * 10^(-7)(2.99792* 10^8)}}

E=930.84 N/C

Therefore answer is 930.84 N/C

Final answer:

To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.

Explanation:

To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.

Subbing in the given values, we can compute Em as:

Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]

This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.

Learn more about Electromagnetic Waves here:

brainly.com/question/29774932

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Somebody tell me what the answer is please

Answers

Answer:

The acceleration of the ball is 200 m/s^2

Answer:

A

Explanation:

200 m/s squared

find the area of a right triangle with sides 15.0 cm, 20.0 cm, and 25.0 cm. express your answer with the correct number of significant figures

Answers

The area of this triangle can be calculated using herons formula since th three sides are given. It is expressed as:

A=sqrt( s(s-a)(s-b)(s-c))

where s is equal to a+b+c / 2

s=15 +20 +25 /2=30
A=sqrt( 30(30-15)(30-20)(30-25))
A= 150 cm^3

A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?

Answers

Answer:

a) 10.29° upstream

b) t=338.7s

Explanation:

If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

\alpha =atan((0.5)/(1))=26.56°

D=√(1^2+0.5^2)=1.118km

The realitve velocity of the boat respect to the water is:

V_(B/W)=[3*cos\beta ,3*sin\beta ]  where β is the angle it has to be pointed at.

From the relative mvement equations:

V_(B/W)=V_B-V_W  where V_B=[V*cos\alpha ,-V*sin\alpha ]

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

(3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2

V^2-4*V*sin\alpha -5=0  Solving for V:

V = 3.3m/s   and   V=-1.514m/s   Replacing this value into one of our previous x or y-axis equations:

\beta =acos((V*cos\alpha )/(3) ) = 10.29°

The amount of time:

t = D/V = (1118m)/(3.3m/s) =338.7s

What is a light year​

Answers

Answer:

A light-year is the distance light travels in one year.

Answer:

Explanation:

a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 million million miles).

Explain how the Doppler effect works for sound waves and give some familiar examples.

Answers

Answer and Explanation:

Doppler effect : According to Doppler effect whenever there is a relative motion between the source and observer then there is an increase or decrease in frequency of sound light or waves.

REASON OF DOPPLER EFFECT : Doppler effect is mainly due to the sudden change in pitch of the sound

EXAMPLE OF DOPPLER EFFECT : The best example of Doppler effect is when an ambulance passes and when it comes closer then the frequency of the siren increases and when it goes away its frequency decreases.