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kherson [118]
3 years ago
7

A torque of 50 Nm is applied to a grinding wheel with moment of inertia 20 kg m2 for 10 s. If it starts from rest, what is the a

ngular velocity of the wheel after the torque is removed?
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
8 0

Answer:

Final angular velocity, \omega_f=25\ rad/s

Explanation:

Given that,

Torque, \tau=50\ N-m

Moment of inertia, I=20\ kg m^2

Time, t = 10 s

Initial angular velocity, \omega_i=0

The relation between the torque and the moment of inertia is :

\tau=I\times \alpha

\alpha is the angular acceleration

\alpha =\dfrac{\tau}{I}

\alpha =\dfrac{50}{20}

\alpha =2.5\ rad/s^2

Using first equation of kinematics to find the final angular velocity of the wheel. It is given by :

\omega_f=\omega_i+\alpha t

\omega_f=0+2.5\times 10

\omega_f=25\ rad/s

So, the final angular velocity of the wheel is 25 rad/s. Hence, this is the required solution.

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Y_Kistochka [10]
B.False

Einstein's vision of GR is NOT that somehow Gravity comes along and alters (indents?) some existing structure.
It is that Gravity (with its four possible sources) actually determines the entire global structure of Space-Time in which such sources are extant.
6 0
3 years ago
A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when
vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}

Then F = ma = 0.04 x 2500 = 100N

5 0
3 years ago
Help !!
Vaselesa [24]
The answer is ; 6cm

Hope this helps!
Please give Brainliest!

This is because of the diagram below:

6 0
2 years ago
WILL MARK BRAINLIEST
ANEK [815]

Answer:

Pressure= force/ area

12/ 2033.33= 24400 N/m2.

i believe this is it :)

3 0
3 years ago
an open tank has the shape of a right circular cone (see figure). the tank is 8 feet across the top and 6 feet high. how much wo
Liono4ka [1.6K]

The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.

Given that, the tank is 8 feet across the top and 6 feet high

By the property of similar triangles, 4/6 = r/y

6r = 4y

r = 4/6*y = 2/3*y

Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²

The weight of each disc is m = ρw* A

m = 62.4* 4π/9*y² = 87.08*y²

The distance pumped is 6-y.

The work done in pumping the tank by pumping the water over the top edge is

W = 87.08 ∫(6-y)y² dy

W = 87.08 ∫(6y³ - y²) dy

W =  87.08 [6y⁴/4 - y³/3]

W =  87.08 [3y⁴/2- y³/3]

The limits are from 0 to 6.

W =  87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs

The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.

To know more about work done:

brainly.com/question/16650139

#SPJ4

7 0
1 year ago
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