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Mandarinka [93]
3 years ago
14

A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin

ation is
Physics
1 answer:
sveticcg [70]3 years ago
7 0

Answer:

<em>The equivalent resistance of the combination is R/100</em>

Explanation:

<u>Parallel Connection of Resistances</u>

If resistances R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

It's known the 10 parts or resistance R/10 were connected in parallel, thus the electric resistance is:

\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}+...+\frac{1}{R/10}

Note the sum consists of 10 equal terms. Operating on each term:

\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}

The sum of 10 identical fractions yields 10 times each fraction:

\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}

Solving for Re needs to take the reciprocal of both sides of the equation:

R_e=R/100

The equivalent resistance of the combination is R/100

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Answer:

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Explanation:

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The Atmosphere in Jupiter is highly violent and thinner as compared with Saturn.

The layers of atmosphere on Saturn is thicker because of lower gravity of Saturn and hence they are calmer.  

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To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug
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To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

\tau = Fr

Where,

F = Force

r = Radius

Replacing we have that,

\tau = Fr

\tau = 21cm (\frac{1m}{100cm})* 550N

\tau = 11.55Nm

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2

I = 0.394kg\cdot m^2

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}

\alpha = \frac{11.55}{0.394}

\alpha = 29.3 rad/s^2

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians (\pi / 3), therefore

W = \tau \theta

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Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?
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Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

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Read 2 more answers
A worker pushed a 33 kg block 6.1 m along a level floor at constant speed with a force directed 23° below the horizontal. if the
jenyasd209 [6]
The work done occurs only in the direction the block was moved - horizontally. Work is given by:

W = F(h) * d

Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.

Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):

F(h) = F(app)cos(23)

Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:

N = mg + F(v) = mg + F(app)sin(23)

Now we can get down to business and solve for F(app) - as mentioned above:

F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8

Now that we have F(app), we can find the exact value of F(h):

F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7

And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3

Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
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3 years ago
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