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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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3 years ago
A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s
insens350 [35]

Answer:

24.531 m

Explanation:

t = Time taken = 1.7 s

u = Initial velocity = 6.1 m/s

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=6.1\times 1.7+\dfrac{1}{2}\times 9.8\times 1.7^2\\\Rightarrow s=24.531\ m

The initial height of the rock above the ground is 24.531 m

7 0
2 years ago
7. En la sala de una casa hay una gran ventana de vidrio, por la que se presenta una pérdida significativa de calor; las medidas
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7 0
3 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

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5 0
3 years ago
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