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mrs_skeptik [129]
3 years ago
10

True or false: You can calculate a three variable equation knowing only one of the variables.​

Physics
1 answer:
makvit [3.9K]3 years ago
3 0

Answer:

false

Explanation:

you cant calculate a three variable equation knowing only one of the variables

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Neurons in our bodies carry weak currents that produce detectable magnetic fields. A technique called magnetoencephalography, or
Stolb23 [73]

Answer:

I = (1.80 × 10⁻¹⁰) A

Explanation:

From Biot Savart's law, the magnetic field formula is given as

B = (μ₀I)/(2πr)

B = magnetic field = (1.0 × 10⁻¹⁵) T

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

r = 3.6 cm = 0.036 m

(1.0 × 10⁻¹⁵) = (4π × 10⁻⁷ × I)/(2π × 0.036)

4π × 10⁻⁷ × I = 1.0 × 10⁻¹⁵ × 2π × 0.036

I = (1.80 × 10⁻¹⁰) A

Hope this Helps!!!

3 0
3 years ago
A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

3 0
3 years ago
The cable holding a 2125 kg elevator has a maximum strength of 21,750 N. What is the maximum upward acceleration the cable can g
vivado [14]

Answer:

10.23m/s^2

Explanation:

GIven data

mass of elevator = 2125 kg

Force= 21,750 N

Required

The maximum acceleration upward

F= ma

a= F/m

a=21,750/2125

a= 10.23m/s^2

Hence the acceleration is 10.23m/s^2

4 0
3 years ago
A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
A red blood cell contains 4.8 107 free electrons. What is the total charge of these electrons in the red blood cell?
tatuchka [14]

Answer:

Charge, q=7.68\times 10^{-12}\ C

Explanation:

It is given that,

The number of electron in a RBCs, n=4.8\times 10^7

We need to find the total charge of these electrons in the red blood cell. Let it is q. Using the quantization of charge as follows :

q = ne

e is the change on electron

q=4.8\times 10^7\times 1.6\times 10^{-19}\\\\q=7.68\times 10^{-12}\ C

So, the net charge is 7.68\times 10^{-12}\ C.

7 0
3 years ago
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