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3 years ago

The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:

1 answer:

7 0

To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and $w1 = 1 revolution/sec = 2Pi (rad/s)$
Now we can go to proportion
$v1=v2$
$w1*r1 = w2r2$ $w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)$
$w2 = w3 (which is the angular velocity of the rear wheel) 
$
SOLVING FOR A : $v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec$
$v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)$ $= 9.99$ or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!

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B) 18.5 m/

s west

Explanation:

Scalar quantity has MAGNITUDE only

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3 years ago

Boat A and Boat B have the same mass. Boat A’s velocity is three times greater than that of Boat B. Compared to the kinetic ener

Helen [10]

Answer:

Explanation:

D. Nine times as much.

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3 years ago

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2 years ago

The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the ce

NISA [10]

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

$V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}$ -------equation (1)

The electric field on the sphere is also given as;

$E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}$

$|q |= 4 \pi \epsilon _o} R^2E$

Substitute in the value of q in equation (1)

$V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V$

Therefore, the potential at the center of the sphere is -924 V

8 0

3 years ago

The sound intensity a distance d1 = 15.0 m from a chain saw is 0.260 W/m2. What is the sound intensity a distance d2 = 27.0 m fr

bagirrra123 [75]

Answer:

0.08024 W/m²

Explanation:

P = Power

$d_1$ = Distance = 15 m

$I_1$ = Intensity at 15 m = 0.26 W/m²

$d_2$ = Distance = 27 m

$I_1$ = Intensity at 27 m

Sound intensity is given by

$I=\frac{P}{4\pi d^2}$

$\\\Rightarrow I\propto \frac{1}{d^2}$

So, we have the relation

$\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}\\\Rightarrow I_2=\frac{d_1^2\times I_1}{d_2^2}\\\Rightarrow I_2=\frac{15^2\times 0.26}{27^2}\\\Rightarrow I_2=0.08024\ W/m^2$

The sound intensity at the given distance is 0.08024 W/m²

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3 years ago