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maksim [4K]
2 years ago
15

A motorcycle has a magnet attached to the rim of its front wheel. The front tire has a diameter of 60 cm. A magnetic pickup is a

ttached to the front forks and receives an electronic pulse each time the magnet passes by. How far has the motorcycle traveled when the pickup has received 10 pulses?
Physics
1 answer:
pashok25 [27]2 years ago
8 0

Answer: 1.9m

Explanation:

60 cm x 22/7 x 10cm = 1.9m

You might be interested in
Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo
nalin [4]

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

3 0
2 years ago
A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are t
mel-nik [20]

If the book is placed in the middle, the forces acting on <em>p</em> and <em>q</em> is 5N. If the book is moved 50 cm from <em>q</em>, the forces at <em>p</em> and <em>q</em> can be solved by doing a moment balance

<u>With </u><u><em>p</em></u><u> as the pivot;</u>

Fq (2 m) = 10 N (0.5 m)

Fq = 2.5 N

Fp = 10 N - 2.5 N = 7.5 N

5 0
2 years ago
A vessel of 0.25 m3 capacity is filled with saturated steam at 1500 kPa. If the vessel is cooled until 25% of the steam has cond
soldier1979 [14.2K]

Answer:

the final pressure is P=353.5 kPa and the heat lost by the system (heat transferred) is Q= -3614.7327 kJ ( negative sign means heat outflow)

Explanation:

since the steam is at saturated state at 1500 kPa , from tables of saturated steam  

at P= 1500 kPa → specific volume v= 0.13225 m³/kg , ug=2593.3 kJ/kg

thus

m = V/v = 0.25 m³ / 0.13225 m³/kg = 1.89 kg

for condensation until 25% of steam, then mass of steam is

mg₂ = 0.25* 1.89 kg = 0.4725 kg

if we neglect the volume occupied by the liquid , then the steam occupies

v₂ = V/m = 0.25 m³ / 0.4725 kg =0.529 m³/kg

returning to the saturated steam table

at v₂=0.529 m³/kg →  353.5 kPa , uL₂= 584.1585 kJ/kg , ug₂= 2548.4965 kJ/kg

then from the first law of thermodynamics

ΔU= Q - W , but since V=constant , dV=0 and W=∫PdV =0

Q= ΔU

Q= (mg₂*ug₂+ml₂*uL₂) - (m*ug) = 1.89 kg* (0.25*2548.4965 kJ/kg + 0.75*584.1585 kJ/kg - 2593.3 kJ/kg) = -3614.7327 kJ

5 0
2 years ago
The wind blows a jay bird south with a force of 300 Newtons. The
adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The  jay bird flies north, against the wind, with a force of 100  newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

F=\sqrt{300^2+100^2} \\\\F=316.22\ N

Hence, the net force on the jay bird is 316.22 N.

6 0
2 years ago
Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
kobusy [5.1K]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere  ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon  }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Q₁ = 572.8 * 10^{-9} C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2}  = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  =0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
2 years ago
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