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2 years ago

The motor of a weed trimmer spins

Physics

1 answer:

tester [92]2 years ago

7 0

The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A

<h3>What is a weed trimmer?</h3>

The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.

The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.

Learn more about weed trimmer:brainly.com/question/23160904

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A uniform electric field of magnitude 4.6 ✕ 104 N/C is perpendicular to a square sheet with sides 3.0 m long. What is the electr

rodikova [14]

Answer:

41.4* 10^4 N.m^2/C

Explanation:

given:

E= 4.6 * 10^4 N/C

electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field

then electric flux = ∫ E*n dA

= ∫ 4.6 * 10^4 * 3*3

= 41.4* 10^4 N.m^2/C

3 0

3 years ago

Read 2 more answers

N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2

liberstina [14]

Answer:

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

5 0

3 years ago

an aircraft landing on an air craft carrier is brought to a complete stop from an inital velocity of 215km/hr in 2.7 seconds. wh

worty [1.4K]

u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2

Hope it helps

8 0

2 years ago

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$