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1 year ago

A porter can carry 40 bricks of 10 N load of each. He can carry up to 75m in 40 sec, calculate his power.

Physics

1 answer:

alexira [117]1 year ago

4 0

Answer:

750W

Explanation:

40×10= 400N

work done= force × distance

=400 × 75

=30000 J

Power= work done/ time

= 30000 ÷ 40

= 750 W

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Determine the rotational inertia of the construction about an axis perpendicular to the picture and passing through point B.

sasho [114]

Answer:

m* 8.0 kg*m²

Explanation:

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In the attached picture, neglecting the masses of the rods joining the masses, and taking into account that all masses are equal each other, being equal to m kg, we can write the following equation:

I = m₁*r₁² + m₂*r₂² + m₃*r₃² + m₄*r₄²,

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3 years ago

What must occur for work to be done on an object?

stiv31 [10]

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andrew-mc [135]

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3 years ago

A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes f

serg [7]

Answer:

19080667.0818 m/s

0.637294 m

$2.1875\times 10^{15}$

Explanation:

m = Mass of deuterons = $3.34\times 10^{-27}\ kg$

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s$

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

$\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m$

The diameter is 0.637294 m

Current is given by

$i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}$