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PSYCHO15rus [73]

3 years ago

12

Why does a car slide when it rolls over an icy patch on a road?

1 answer:

Rainbow [258]3 years ago

7 0

Hello. The answer, though I don't know what options you need to be answered, is that there is less friction between the car tires and the ice, than when they are on the road. Causing the car to slide.

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A tall tube is evacuated and its stopcock is closed. The open end of the tube is immersed in a container of water (density 10^3

LuckyWell [14K]

Answer:

10.19 m

Explanation:

Water will rise to equalize the pressure inside and outside the tube.

The equation of pressure is given by

$p=\rho gh$

Where,

p = Pressure of air = 10⁵ N/m²

ρ = Density of water = 10³ kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Height the water will rise

$h=\frac{p}{gh}\\\Rightarrow h=\frac{10^5}{9.81\times 10^3}\\\Rightarrow h=10.19\ m$

∴ The water will rise by 10.19 m.

7 0

2 years ago

An electromagnetic wave is produced by:

Ray Of Light [21]

A. an accelerating charged charged particle or changing magnetic fields

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=what%20%5C%3A%20is%20%5C%3A%20refraction%20%5C%3A%20of%20%5C%3A%20light%20%5C%3A%20%20%7B%3F%7

topjm [15]

When light passes through one transparent medium to another transperent medium it bends and that beding of light is know as refraction of light !

8 0

2 years ago

Read 2 more answers

What is the equation that relate electric potential (voltage) to electric field?

erma4kov [3.2K]

V= I x R
I= V / R
r= V / I

7 0

2 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago