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Artist 52 [7]
3 years ago
6

A 60-V potential different is applied across a parallel combination of a 10-ohm and 20-ohm resistor. What is the current in the

10-ohm resistor? Show your work.
Physics
2 answers:
galina1969 [7]3 years ago
8 0

Answer:

(straight out of physics book)

Explanation:

Beats are two tones of slightly different frequency sounded together.  When two slightly mismatched forks are sounded together, beats are heard. Beating in waves is the result of "superposing" - sort of adding together - two waves of different but similar frequencies - eg 50 Hz and 51Hz. The waves go in and out of phase as a sort of continuous or running interference pattern. In the case of sound, you'd hear it as a continuous change in the sound intensity loud in phase, soft out of phase. An example is walking side by side to another person. Sometimes you two will be in the same step and other times you’ll be in different step. If you step 60 times in a minute and your friend takes 63 in that minute, your friend gains three steps per minute on you.  This shows that you two will be in step three times per minute. This goes for forks and frequency as well. When one fork vibrates 242 times per second  and the other 240 per second, they are in step twice each second. This results in 2 hertz beat frequency.

kodGreya [7K]3 years ago
3 0
V=IR
60-V
The current that passes through a 10-ohm resistor = I
I=60/10
6 amperes
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You push a shopping cart for fun with a force of 60N. If the shopping cart has a mass of 12kg, what is the
posledela

Answer:

\huge\boxed{\sf a = 5 \ ms^{-2}}

Explanation:

<u>Given:</u>

Force = f = 60 N

Mass = m = 12 kg

<u>Required:</u>

Acceleration = a = ?

<u>Formula:</u>

F = ma

<u>Solution:</u>

Rearranging formula

a = F / m

a = 60 / 12

a = 5 ms⁻²

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>
4 0
3 years ago
A shopper pushes a 5.32 kg grocery cart
Juli2301 [7.4K]

Answer:

\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

\text { Acceleration }=\frac{\text { force }}{\text { mass }}

Given that,

Mass = 5.32 kg

\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}

x=-28.7^{\circ}

F = 12.7N

Normal force = mg + F sinx,  

“m” being the object's "mass",  

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}

\text { Acceleration }=\frac{58.232}{5.32}

\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}

\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}

7 0
3 years ago
Read 2 more answers
List two applications of electrostatics.
Tems11 [23]
Atmospheric electricity and storms,electric current in a vacuum,spark discharge,electrostatic control filters and industrial electrostatic  separation <- those are just a few
4 0
3 years ago
Types of cell combination? in electricity​
Grace [21]

Answer:

Series and Parallel

7 0
3 years ago
A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire
Morgarella [4.7K]

Answer:

electric field in the wide wire is

E₂ =\frac{E}{4}

Explanation:

given

length of the copper wire = L

radius of the copper wire r₁ = b

length of the second copper wire = L

radius of the second copper wire r₂ = 2b

electric field in the narrow wire = E₁=E

recall

resistance R = ρL/A

where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.

Resistance of narrow wire, R₁

R₁ = ρL/A

where A  = πb²

R₁ = ρL/πb²---------- eqn 1

Resistance of wide wire, R₂

R₂ = ρL/A

where A = π(2b)²

R₂ = ρL/π(2b)²

R₂ = ρL/4πb²-------------- eqn 2

R₂ = ¹/₄(ρL/πb²)

comparing eqn 1 and 2

R₁ = 4R₂

calculating the current in the wire,

I = E/(R₁ + R₂)

recall

R₁ = 4R₂

∴ I = E/(4R₂ + R₂)

I = E/5R₂

calculating the potential difference across R₁ & R₂

V₁ = IR₁

I = E/5R₂

∴ V₁ = ER₁/5R₂

R₁ = 4R₂

V₁ = 4ER₂/5R₂

∴V₁  = ⁴/₅E

potential difference for R₂

V₂= IR₂

I = E/5R₂

∴ V₂ = ER₂/5R₂

V₂ = ER₂/5R₂

∴V₂  = ¹/₅E

so, electric field E = V/L

for narrow wire E₁ = V₁/L ----------- eqn 3

for wide wire, E₂ = V₂/L------------ eqn 4

compare eqn 3 and 4

E₂/E₁ = V₂/V₁( L is constant)

E₂/E₁ = ¹/₅E/⁴/₅E

E₂ = E₁/4

note E₁ = E

∴E₂ =\frac{E}{4}

8 0
4 years ago
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