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elena-14-01-66 [18.8K]

2 years ago

11

gabriel and arely have performed an experiment and are not sure their results are valid. what should they do to check their resu

lts

2 answers:

zubka84 [21]2 years ago

7 0

Do the math and then compare your results to theirs.

Arisa [49]2 years ago

6 0

Yeah, or they can compare the results to other people

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Three multiple choice science questions.

Ratling [72]

C
A
B
I HOPE THIS HELPS

6 0

3 years ago

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

Greeley [361]

Answer:

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

Explanation:

<u>Electric Resistance</u>

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.

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

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

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

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

This sum is repeated 10 times. Operating each term:

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

All the terms have the same denominator, thus:

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

Taking the reciprocals:

$R_e=R/100$

The equivalent resistance of the combination is R/100

6 0

3 years ago

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0

3 years ago

Read 2 more answers

Kepler deduced this law of motion from observations of Mars. What information confirms his conclusion that the orbit of Mars is

Leno4ka [110]

Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.

This led to Kepler's law of orbital motion.

First Law: Planetary orbits are elliptical with the sun at a focus.

Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.

Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.

It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.

6 0

3 years ago

In the Millikan oil drop experiment the measured charge of any single droplet was always a whole number multiple of -1.60 x 10-1

BaLLatris [955]

Answer:

Number of electrons, n = 6

Explanation:

Total charge in a single droplet, $q=-9.6\times 10^{-19}\ C$

The measured charge of any single droplet, $e=-1.6\times 10^{-19}\ C$

Let n is the number of excess electrons are contained within the drop. According to the quantization of charge :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{-9.6\times 10^{-19}}{-1.6\times 10^{-19}}$

n = 6

So, there are 6 electrons contained within the drop. Hence, this is the required solution.

7 0

3 years ago