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3 years ago

Which best describes the motion of the object between 1

Physics

1 answer:

ratelena [41]3 years ago

6 0

Answer:

the object has negative acceleration and eventually

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On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

aleksklad [387]

Ey64udnhsufifjyueuhhekr

8 0

3 years ago

If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre

Dmitrij [34]

M represent mass in kg
v represents speed in m/s
r represents radius in m

Now, just substitute these into the formula:
 $F = \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }$

3 0

3 years ago

Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an

vampirchik [111]

Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

V = I×B/qnd

Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

I = current =?,

n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

q = magnitude of an electronic charge=1.609×10^-19c

B = strength of magnetic field = 5T

d = thickness of sensor = 0.8mm = 0.0008m

By slotting in the parameters, we have that

0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008

0.0015 = I×5/7.446×10^-8

I = (0.0015 × 7.446×10^-8)/5

I = 111.69*10^(-12)

I = 111.69 pA

3 0

3 years ago

A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0

3 years ago

An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w

adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 × $10^{-2}$ m

diameter d = 0.074 cm = 74 × $10^{-5}$ m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 × $10^{-5}$ ×5 × $10^{-2}$

area = 1162.3892 × $10^{-5}$ m²

so here power emitted is express as

power emitted = E × σ × area × (temperature)^4

put here all value

power emitted = 0.300× 5.67 × 1162.3892 × $10^{-5}$ × (3068)^4

power emitted = 1.75 W

5 0

3 years ago