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alex41 [277]
3 years ago
10

An overhang hollow shaft carries a 900 mm diameter pulley, whose centre is 250 mm from the centre of the nearest bearing. The we

ight of the pulley is 600 N and the angle of lap is 180°. The pulley is driven by a motor vertically below it. If permissible tension in the belt is 2650 N and if coefficient of friction between the belt and pulley surface is 0.3, estimate, diameters of shaft, when the internal diameter is 0.6 of the external. Neglect centrifugal tension and assume permissible tensile and shear stresses in the shaft as 84 MPa and 68 MPa respectively.
Physics
1 answer:
ddd [48]3 years ago
6 0

Answer:yu

Explanation:

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3. What will happen to the black bass and blue gill as the floor of the ponds fills with organic
yKpoI14uk [10]

Answer: die

Explanation: oyxagan all goon bc of all dat suffs

8 0
3 years ago
How do resistors in parallel affect the total resistance?
4vir4ik [10]

Answer:

They're going to increase the total resistance as R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

I = I_1 + I_2 + ... + I_N

where

I_i = \frac{V_i}{R_i}

but

V_i = V_j = V for i,j= 1, 2,..., N

so

I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}

where

R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

4 0
3 years ago
A 5.00X10^5 kg rocket is accelerating straight up. Its engines produce 1.250X10^7 N of thrust, and air resistance is 4.50X10^6 N
katrin [286]
<span>6.20 m/s^2 The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be 9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N Add in the atmospheric drag and you get 4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N Now subtract that total drag from the thrust available. 1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So 3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2 Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
8 0
3 years ago
The sensor in the torso of a crash test dummy records the magnitude znd direction of the net force acting on the dummy.If the du
patriot [66]

Answer : The force will be 4501.9

We can see that, two forces acting on the dummy in two different direction.

We know that, here two forces are given in perpendicular direction with each other.

We know the force is the vector addition law so, we will use the Pythagoras theorem for the resultant of the vectors

Now, the net force is

F_{net} = \sqrt{{F_{1}}^{2}+ {F_{2}}^{2}}

Two forces is given,

F_{1} = 130.0 N

F_{2} = 4500.0 N

Now, the net force is

F_{net} = \sqrt{(130)^{2}+ (4500)^{2}}N

F_{net} = \sqrt{20266900}N

F_{net} = 4501.9 N

Hence, the force will 4501.9 N

8 0
3 years ago
Read 2 more answers
Distinguish between refraction,reflection and total internal reflection
Ipatiy [6.2K]

The contrast between total internal reflection and refraction is that total internal reflection is the full underwater build with no defeat of brightness, whereas refraction is the shift in the path of a tide that is parting from one medium to another.

7 0
2 years ago
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