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3 years ago

The rod connected halfway along the 0.660 m radius of a wheel exerts a 2.27 × 10^5 N force. How large is the maximum torque?

Physics

1 answer:

Klio2033 [76]3 years ago

3 0

Answer:

1.498×10⁵ Nm

Explanation:

From the question above,

Torque (T) = Force (F) × radius (r)

T = F×r............................. Equation 1

Where T = Torque, F = force, r = radius .

Given: F = 2.27×10⁵ N, r = 0.660 m

Substitute these values into equation 1

T = 2.27×10⁵×0.660

T = 1.498×10⁵ Nm

Hence the torque of the wheel is 1.498×10⁵ Nm

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boyakko [2]

Answer:

the one with a higher mass

Explanation:

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4 years ago

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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)

SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c² => c = 0.707 m

Charge to the right: In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x $10^{9}$ N m²/C²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below: For y direction

e = kq/r²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

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4 years ago

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