Solve the given parallel circuit by computing the desired quantities.

Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts

Gala2k [10]

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

$v=u+at$

Substitute the values

$10=0+162a=162a$

$a=\frac{10}{162}=0.0617m/s^2$

$s=ut+\frac{1}{2}at^2$

Substitute the values

$s=\frac{1}{2}(0.0617)(162)^2=809.6 m$

$s'=vt'=10\times 822=8220 m$

$a'=\frac{v}{t''}=\frac{10}{246}$

$s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m$

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m

6 0

4 years ago

Read 2 more answers

When an object is in orbit, it is falling at the same rate at which the Earth is curving. Please select the best answer from the

olga2289 [7]

The statement “When an object is in orbit, it is falling at the same rate at which the Earth is curving” is true. The speed of a satellite orbiting the earth depends only on the mass of the earth and the mass of the satellite.

7 0

3 years ago

A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di

BlackZzzverrR [31]

Answer:

$r =3 *\sqrt{2} = 4.24 cm$

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

$E_{MAX}= \frac{KQ}{r^{2}}$

electric field inside the sphere can be determine by using below relation

$\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}$

$r = \sqrt{2}R$

$r =3 *\sqrt{2} = 4.24 cm$

4 0

3 years ago

Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t

ivolga24 [154]

According to newton's law of gravitation, the gravitational force(F) is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

Fg ∝ (Mp)(Mm)/d²

Fg = G(Mp)(Mm)/d²

This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg = G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

which is the value we obtained on earth for acceleration due to gravity.

To know more about gravitational constant, visit:

brainly.com/question/13959861

#SPJ4

5 0

1 year ago

What determines if a star turns into a red giant or a red supergiant?

mixer [17]

Answer:

Explanation:

As stars age, they evolve away from the main sequence to become red giants or supergiants. The core of a red giant is contracting, but the outer layers are expanding as a result of hydrogen fusion in a shell outside the core. The star gets larger, redder, and more luminous as it expands and cools.

3 0

3 years ago