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galben [10]
2 years ago
11

The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.

If the belt is subjected to a force of P=200N, determine the angular velocity of the wheel, three seconds after the force is applied.b. Determine the speed of its center of mass O, three seconds after the force is applied.
Physics
1 answer:
pickupchik [31]2 years ago
3 0

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × 10^{-6} = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

w_{f} = w_{i} + α × t

Where

w_{f} is the final angular velocity

w_{i} is the initial angular velocity

α is the angular acceleration

t is the time taken

w_{i} is 0 (∵ initially it starts from rest)

By substituting the values we get

w_{f} = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

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The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900
Sedaia [141]

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}.

(b) Now horizontally,

L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.

3 0
2 years ago
Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an index of ref
blsea [12.9K]

Answer:

the new diameter of the third ring = 0.607 mm

Explanation:

Consider the radius of \\m ^{th}\\ bright ring when air is in between the lens and the plate ;

r = \sqrt {\frac{(2m+1)\lambda R}{2}}

Using the expression: r (n)= \sqrt {\frac{(2m+1)\lambda R}{2n}} for the radius of the  \\m ^{th}\\ bright ring since the liquid (water ) of the refractive index 'n' is now in between the lens and the plate;

where;

\\m ^{th}\\ = number of fringe

λ = wavelength

R = radius

n = refractive index of water

Now ;

r (n)=\frac {r}{n}}

the radius r of the third bright ring when the air is in between lens and plate = r= \frac{0.700 \ mm}{2} \\\\

r= 0.35 \ mm

The new radius of the third bright fringe is r(3) = \frac{0.35}{\sqrt 1.33}

r(3) = 0.3035 \ mm

Calculating the new diameter ; we have:

d(3) = 2(r(3))

d(3) = 2(0.3035)

d = 0.607 mm

Thus, the new diameter of the third ring = 0.607 mm

7 0
2 years ago
How much does a person weigh if it takes 700 kg*m/s to move them 10 m/s<br><br> NEED ASAP
madreJ [45]

Answer:

\huge\boxed{m = 70 \ kg}

Explanation:

<u>Given Data:</u>

Momentum = P = 700 kg m/s

Velocity = v = 10 m/s

<u>Required:</u>

Mass = m = ?

<u>Formula:</u>

P = mv

<u>Solution:</u>

m = P / v

m = 700 / 10

m = 70 kg

\rule[225]{225}{2}

Hope this helped!

<h3>~AnonymousHelper1807</h3>
5 0
2 years ago
A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
We know that the normal component of Fg is just Fg, which is equal to as 1110N. 
From the geometry, the normal component of Fp can be calculated: 
Fpn = Fp * cos(θp) 
= 1016.31 N * cos(53) 
= 611.63 N 

The total normal force Fn then is: 
Fn = Fg + Fpn 
= 1110 + 611.63
= 1721.63 N</span>

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
= (Fp * sin(θp)) + 0 
= 1016.31 N * sin(53) 
= 811.66 N</span>

 

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0
2 years ago
The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis
lord [1]

Answer:

The distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

Explanation:

As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is <em>equivalent</em> to the distance it travels around the Sun in a single orbit.

<em>The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)</em>

Where,

<em>R = Orbital radius (in this case) = 30.1 AU</em>

<em />

Plug the value of R in the equation (A):

<em>(A) => The circumference of the circle = 2*π*(30.1)</em>

<em> The circumference of the circle = </em><em>60.2π</em>

Therefore, the distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

5 0
2 years ago
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