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kondor19780726 [428]
3 years ago
15

A hall probe serves to measure magnetic field strength. such a probe consists of a poor conductor 0.155 mm thick, whose charge-c

arrier density is 1.05 × 1025 m3. when a 2.67-a current flows through the probe, the hall voltage is measured to be 4.89 mv. what is the magnetic field strength?
Physics
1 answer:
jeka943 years ago
4 0
We are given the following:
thickness, T = 0.155 mm
charge-carrier density, n = 1.05 x 10^25 / m3
current, I = 2.67 A
voltage, V = 4.89 mV

We are asked to get the magnetic strength, B

To solve this, we use the Hall Effect formula:
V = - IB / neT
where e is the charge of an electron which is e = 1.60 x 10^-19 and B is the magnetic field
Rearranging the formula
B = -V n e T / I
Substituting the given values
B = - (4.89 x 10^-3 V) (1.05 x 10^25 / m3) (0.155 x 10^-3 m) / 2.67 A
B = -2.98 x 10^18 Teslas
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4 0
3 years ago
An object is launched upwards at an initial speed of 3 m/s. What is the maximum height reached by the object from where it was l
ollegr [7]

Answer:

the maximum height reached by the object from where it was launched is 0.4591 m

Explanation:

initial speed of the object, u= 3 m/s

The velocity at the maximum height will always be 0.

Therefore,  final velocity, v= 0 m/s  

Using the Newton's  equation of motion,

v^2 - u^2 = 2*g*h(max)

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h(max) = -u^2 /2* g

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g= - 9.8 m/s^2

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h(max) =  0.4591 m

the maximum height reached by the object from where it was launched is 0.4591 m

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4 0
2 years ago
A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.
Maksim231197 [3]

Answer:

v=33.764m/s

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

F_{c}=\frac{mv^{2} }{r}

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

4 0
3 years ago
Light, dry snow is called powder. Skiing on a powder day is different than skiing on a day when the snow is wet and heavy. When
kramer

Answer:

Explanation:

Given:

- Acceleration a_dry = 0.4*a_wet

- initial velocity V_i

Find:

By how much does time taken to stop of dry snow differ from time taken to stop on wet snow?

Solution:

- Time taken to stop on wet snow t_wet:

                            V_f = V_i + a_wet * t_wet

                             t_wet = - a_wet / V_i

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                             t_dry = - a_dry / V_i

- Difference in t_wet and t_dry:

                             t_dry - t_wet =  - a_dry / V_i  + a_wet / V_i

                             dt = (a_wet - a_dry) / V_i

                             dt = (a_wet - 0.4*a_wet) V_i

                             dt = 0.6*a_wet / V_i

3 0
3 years ago
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