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geniusboy [140]

2 years ago

8

A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc

eleration to be 150 rad/s2

1 answer:

kvv77 [185]2 years ago

4 0

Hi there!

We can use the rotational equivalent of Newton's Second Law:

$\huge\boxed{\Sigma \tau = I \alpha}$

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

$\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}$

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erastova [34]

A scientist would write that number as 1.49 x 10⁸ kilometers .

(Or, if the scientist is in France or the UK, he might write it as 1.49 x 10⁸ kilometres .)

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3 years ago

Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre

VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers $d=1.6\ m$

Distance of person from one speaker $x_1=3\ m$

Distance of person from second speaker $x_2=3.5\ m$

Path difference between the waves is given by

$x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}$

for destructive interference m=0 I.e.

$x_2-x_1=\frac{\lambda }{2}$

$3.5-3=\frac{\lambda }{2}$

$\lambda =0.5\times 2$

$\lambda =1\ m$

frequency is given by

$f=\frac{v}{\lambda }$

where $v=velocity\ of\ sound\ (v=343\ m/s)$

$f=\frac{343}{1}=343\ Hz$

For next frequency which will cause destructive interference is

i.e. $m=1$ and $m=2$

$3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda$

$\lambda =\frac{1}{3}\ m$

frequency corresponding to this is

$f_2=\frac{343}{\frac{1}{3}}=1029\ Hz$

for $m=2$

$3.5-3=\frac{5}{2}\cdot \lambda$

$\lambda =\frac{1}{5}\ m$

Frequency corresponding to this wavelength

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3 years ago

What term refers to the part of a spacecraft that is occupied by the crew for takeoff and landing?

Semenov [28]

Command module ✅

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7 0

2 years ago

The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started

bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

$W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J$

The change in electrostatic potential energy $\Delta U$ , of one point charge q is defined as the product of the charge and the potential difference.

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5 0

3 years ago

A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m

anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

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F = mv²/r

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a = v²/r

Substituting the values in the above equation

13.33 = v²/30

v² = 13.33 x 30

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v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

3 0

3 years ago