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bearhunter [10]
2 years ago
11

4. A 1300 kg car makes a turn with a 45 m radius. Coefficient of static

Physics
1 answer:
asambeis [7]2 years ago
3 0

Answer: v = 11 m/s

Explanation:

If we ASSUME that the road is horizontal, not banked

The maximum friction force available is μmg

This will need to supply all of the required centripetal force

mv²/R = μmg

v = √(μgR) = √(0.28(9.8)(45)) = 11.1121...≈ 11 m/s

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A wagon wheel consists of 8 spokes of uniform diameter, each of mass ms and length L cm. The outer ring has a mass mring. What i
just olya [345]

Answer:

The moment of inertial of the wheel,  I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2

Explanation:

Given;

8 spokes of uniform diameter

mass of each spoke, = M_s

length of each spoke, = L

mass of outer ring, = M_r

The moment of inertial of the wheel will be calculated as;

I = 8I_{spoke} + I_{ring}

where;

I_{spoke is the moment of inertia of each spoke

I_{ring is the moment of inertia of the rim

Moment of inertia of each spoke =\frac{1}{3}M_sL^2

Moment of inertial of the wheel

I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2

6 0
2 years ago
The peak intensity of radiation from Star Beta is 350 nm. In what spectral band is this? UV, radio waves, visible light, or infa
Stels [109]
I say uv all the way
4 0
3 years ago
Read 2 more answers
This is a measure of the quantity of matter.
emmainna [20.7K]

The measure of the quantity of matter would be mass. Mass is measured in kilograms. I hope this helped!:)

3 0
3 years ago
Read 2 more answers
2. When the pump removed the air in<br> the bell, the balloon<br> contracted<br> expanded
IRINA_888 [86]

When the pump removed the air in  the bell, the balloon expanded.

<u>Option: B</u>

<u>Explanation:</u>

In order to construct our own environment in the glass jar known as bell jar system, which can be used to explore and consider our larger environment on Earths, for an instance. Here a glass jar that hinges on an airtight rubber basis i.e seals appropriately. At the top of the jar, a bung is connected to it which passed via a metal tube. It has an adjacent flexible tube that goes to a hand vacuum pump and the best hand-powered pump was made with a wine preserver.

When the pump extracts the air from the bell jar, the pressure inside the balloon naturally decreases. The balloon usually has a air pressure around it, which restricts its size, but when this air is extracted and the pressure around it decreases the gas in the balloon will expand and the balloon seems to be inflating. When you release the air back into the bell jar, it will once again compress back to its actual size.

8 0
3 years ago
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
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