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3 years ago

A wheel rotates clockwise 10 times per second. what is its angular speed answer

1 answer:

3 0

The frequency of the wheel is the number of revolutions per second:
$f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz$
And now we can calculate the angular speed, which is given by:
$\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s$ in the clockwise direction.

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Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2

Mila [183]

Answer:

The angular acceleration is 11.66 rad/s²

Explanation:

Step 1: Given data

Three forces are applied to a solid cylinder of mass 12 kg

F1 = 15 N

F2 = 24 N

F3 = 19 N

R2 = 0.22m

R3 = 0.10m

Step 2: Find the magnitude of the angular acceleration

I = ½mr² = ½ * 12kg * (0.22m)² = 0.29 kg*m²

torque τ = I*α

τ = F2*R2 - F1*R1 = 24N*0.22m - 19N*0.10m = 3.38 N*m

This means

I = ½mr² = 0.29 kg*m²

τ = I*α = 3.38 N*m

0.29 kg*m² * α = 3.38 N*m

α = 11.655 rad/s² ≈11.66 rad/s²

The angular acceleration is 11.66 rad/s²

6 0

3 years ago

How is speed determined?

enot [183]

Speed equals distance divided by time (Speed=Distance/Time).

For example if it takes someone three minutes to go three miles their speed is one mile a minute.

7 0

3 years ago

Lực hút trái đất là dì

nataly862011 [7]

Lực hút trái đất gọi Là trọng lực

5 0

2 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

______ is the percent ionization of a weak acid in water increases as the concentration of acid decreases.

Nuetrik [128]

The ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

Explanation:

Percent ionization is used for quantifying the number of ions present in the weak acid when dissolved in a solution. So it is similar to the pKa value. The percent ionization value can be determined as negative log of dissociation constant. Also the as the number of ions increases in weak acid, the concentration of acid will be decreasing . It can be calculated using the formula for percent ionization as follows:

$Percent ionization = \frac{Ionized acid concentration}{Initial concentration of acid} * 100$

As the water volume or concentration increases, the acid will get diluted much more thus leading to decrease in the concentration of acid.

So the ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

7 0

2 years ago