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FinnZ [79.3K]
3 years ago
14

A motorbike accelerates from rest at 12m/s in 3s on a level road. If the force is 3000n what is the total mass of the cyclist an

d the motorbike?
Physics
2 answers:
AnnyKZ [126]3 years ago
5 0
A( Acceleration)= 12m/s
F(Force)= 3000N
M(Mass)=?

You Know, F= MA (Force= Mass x Acceleration)
then M= ?

stira [4]3 years ago
3 0
We know that:
a=(v-u)/t
a=(12m/s-0)/3s
a=4m/s²
Now:
F=ma
3000N=m*4m/s²
3000N/4m/s²=m
m=750kg
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A very long straight current-carrying wire produces a magnetic field of 25 µT at a distance d from the wire. How far will the ma
daser333 [38]

The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.

B ∝ 1/d

B = magnetic field strength, d = distance from wire

Calculate the scaling factor for d required to change B from 25μT to 2.8μT:

2.8μT/25μT = 1/k

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Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

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\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

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3 years ago
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Answer:

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Explanation:

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