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2 years ago

The circumference of a bicycle wheel is 2 meters. if it rotates at 1 revolution per second then its linear speed is

1 answer:

6 0

Given:
Circumference = 2 m
Angular speed, ω = 1 rev/s = 2π radians/s

If the radius is r, then
2πr = 2
r = 1/π m

The linear (tangential) speed is
v = rω
= (1/π m)*(2π rad/s) = 0.5 m/s

Answer: 0.5 m/s

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Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit

anygoal [31]

Answer:

0.03 A

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

From ohm's law, the voltage, current and resistor are related by the following formula:

Voltage = current × resistor

V = IR

With the above formula, we can obtain the current in the circuit as follow:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

V = IR

12 = I × 470

Divide both side by 470

I = 12 / 470

I = 0.03 A

Thus, the current in the circuit is 0.03 A

3 0

2 years ago

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Which one of the following accurately describes the force of gravity?

Aloiza [94]

B. The gravity acceleration is in the same direction as the force of gravity, and thus towards the centre of the earth

4 0

2 years ago

Identify the volume units that are greater than one liter

Citrus2011 [14]

There are a lot of volume units, most specifically in English units, that are greater than one liter. The following are as follows:

gallon, which is equal to 4.54 liters
minim
barrel
cord
peck
bushel and;
hogshead

Also included are metric units which are dekaliter onwards.

5 0

2 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

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Mama L [17]

Answer:

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5 0

2 years ago

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