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Ierofanga [76]
3 years ago
8

. An oatmeal cookie is dropped on the floor. Is this an inelastic collision?

Physics
1 answer:
DENIUS [597]3 years ago
4 0

Answer:

It is not inelastic because the cookie and floor do not move away together as a unit.

Explanation:

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sertanlavr [38]
A substance made up of only one type of element
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What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

5 0
3 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a
USPshnik [31]
Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
   = (260 N)*(0.33 m)
   = 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
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Answer: 105.93 kg-m²

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Two Force one of 12 Newton and another 24 Newton acts at 90 degree with each other find the resultant of two force and its direc
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Answer:

Fr = 26.83 [N]

Explanation:

To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.

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3 years ago
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kiruha [24]
He will swim 180m if he stays at a constant rate of 30m/min

Explanation:
First I’m going to use 6 min for the 5min 60s
Then if he swims 30m/min then I’m 6min, 30•6 he’ll swim 180m
Hope this helps!
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2 years ago
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