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prisoha [69]
3 years ago
14

If the efficiency and mechanical advantage of a certain machine are given as 65 % and 3 respectively.What is the velocity ratio

of the machine?
a.3.5 %
b.4.6 %
c.7.9 %
d.11.2 %
Physics
1 answer:
Vinvika [58]3 years ago
8 0

Answer:

b. 4.6 %

Explanation:

From the question,

E = M.A/V.R................ Equation 1

Where E = percentage Efficiency of the machine, M.A = machanical accurancy of the machine, V.R = Velocity ratio of the machine

Make V.R the subject of the equation

V.R = M.A/E

Given: M.A = 3, E = 65% = 0.65

Substitute this values into equation 2

V.R = 3/0.65

V.R = 4.6

Hence the right option is b. 4.6 $

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ludmilkaskok [199]

The quantity with units of length only is \sqrt{\frac{hG}{c^3}}

Explanation:

We have to combine the following constants:

- h, Planck constant, with units [m^2][kg][s^{-1}]

- G, the Newton's gravitational constant, with units [m^3][kg^{-1}][s^{-2}]

- c, the speed of light, with units [m][s^{-1}]

The combination of these constant should have units of length only, so with meters (m).

First, we notice that h has [kg] in its units, while G has [kg^{-1}] in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by c^3, so that the new units are:

\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]

Learn more about gravitational constant:

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3 years ago
2. Which of the following is an example of work being done on an object? A prism scatters ultraviolet light into visible light.
Liula [17]
A man pushes a couch across the room is the answer!

7 0
3 years ago
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3 0
4 years ago
A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =
slega [8]

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:  

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:  

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):  

v=∫δv/δy(dy)=U*y/4cx^1/2

 =y/x*(U*y/4cx^1/2)

 =u*y/4x

which is exactly what we needed to demonstrate.  

Also, using u = U*y/δ in the last equation we can obtain:  

v/U=u*y/4*U*x

     =y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

                =δ/4x

                =2.5 * 10^-3

7 0
3 years ago
A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje
kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0
3 years ago
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