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cestrela7 [59]
3 years ago
14

The vertical component of the magnetic induction in the Earth's magnetic field at Hobart is approximately 6×10-5T upward. What e

lectric field is set up in a car travelling on a level surface at 100 km h-1due to this magnetic field? Which side or end of the car is positively charged? Approximately what p.d. is created across a car of typical size?
Physics
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

Explanation:

Magnetic field B = 6 X 10⁻⁵ T.

Width of car = L (Let )

Velocity of car v  = 100 km/h

= 27.78 m /s

induced emf across the body ( width )  of the car

= BLv

= 6 X 10⁻⁵ L X 27.78

166.68 X 10⁻⁵ L

Induced electric field across the width

= emf induced / L

E =  166.68 X 10⁻⁵ N/C

We suppose breadth of a typical car = 1.5 m

potential difference induced

= 166.68 x 1.5 x 10⁻⁵

250 x 10⁻⁵ V

= 2.5 milli volt.

The side of the car which is positively charged depends on the direction in which car is moving , whether it is moving towards the north or south.

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3 years ago
A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. The balls un-de
sladkih [1.3K]

Answer:7.93 m/s

Explanation:

Given

mass of ball m_1=20\ gm

Mass of hanging ball m_2=100\ gm

Length of string L=1\ m

Maximum angle turned \theta _{max}=50^{\circ}

v_o is the initial velocity of ball 1  and 0 is the initial  velocity of ball 2

For Perfectly elastic final velocity of ball 1 and 2 is given by

v_2'=\frac{2m_1}{m_1+m_2}\cdot v_1-\frac{m_1-m_2}{m_1+m_2}\cdot v_2

v_1'=\frac{m_1-m_2}{m_1+m_2}\cdot v_1+\frac{2m_2}{m_1+m_2}\cdot v_2

where v_1and v_2 are the velocity of 1 and 2 before collision

thus v_2'=\frac{2\times 20}{120}v_0-0

v_2'=\frac{v_o}{3}

v_1'=\frac{20-100}{120}\times v_o+0

v_1'=-\frac{2}{3}v_o

By energy conservation on second ball we get

Kinetic energy=Potential Energy

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v_2'=\sqrt{2gl(1-\cos \theta )}

v_2'=\sqrt{2\times 9.8\times 1(1-0.642)}

v_2'=2.64\ m/s

thus v_o=3\times v_2'=7.93\ m/s

5 0
3 years ago
An ultrasound unit is being used to measure a patient's heartbeat by combining the emitted 2.0 MHz signal with the sound waves r
alexandr402 [8]
Hi there, 
for this question we have:
Signal 2.0 MHz = Emitted so we can call it f_e
and we need the Reflected = f_{r}
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted). 
we also know: ΔBeat frequency(max) = 560 Hz = f_{b}
so we have: 
f_{e} - f_{r} = f_{b}
so frequency of Reflected is: 
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz = f_{r}
now you know that Lambda = v/f 
so if we find the lambda with our Emitted then we can find v with the Reflected: 
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=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s 
so the v_{max} is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day

8 0
4 years ago
Which of the following is NOT a symptom associated with hypertension?
zepelin [54]
I would say a or d hope this helps haha ( also if you could do brainliest for this that would be AMASINF bc I really want to rank up, u don’t have to tho)
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Answer:

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Explanation:

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