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2 years ago

14

an electric current 0.75 a passes through a circuit that has a resistance of 175. according to ohm's law, what is the voltage of

the circuit

1 answer:

Aleksandr [31]2 years ago

8 0

Answer:

131.25

Explanation:

i worked it out on a diffrent sheet so its hard to explain

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Please, can somebody help me with this project? I'll give brainlest for the best answer! (Do not answer if you don't know or onl

Sauron [17]

The density of the nickel was greater than that of the quarter and penny, thus, the results supports the hypothesis.

<h3>What is density of substance?</h3>

The density of a substance is a measure of how tightly-packed the particles of the substance are.

Density is calculated as the ratio of the mass of the substance and the volume of the substance.

Density = mass/volume

The hypothesis of the lab to compare the densities of a penny, a nickel, and a quarter is:

If the nickel has a greater density than the quarter and penny, then it will have a greater mass to volume ratio. If the nickel has a lower density than the quarter and penny, then it will have a lower mass-to-volume ratio.

The average mass and the average volume of a penny, a nickel, and a quarter are then used to determine the density of each coin.

Based on obtained results, it would be found that the density of the nickel was greater than that of the quarter and penny. Therefore, the results supports the hypothesis.

In conclusion, the density of a substance depends on the mass and the volume.

Learn more about density at: brainly.com/question/1354972

#SPJ1

7 0

1 year ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

<em></em>

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

3 0

3 years ago

A car travels at a constant velocity of 20.0 meters/second for 15 seconds. What is the power of the car if the initial force app

Zielflug [23.3K]

Work done by the force = Force x displacement. Power = work done/time = F.s/t = F.u.t/t = F.u = 95 x 20 = 1900J. {S=ut because acceleration is zero since car is moving at constant velocity}.

3 0

3 years ago

Read 2 more answers

The ability of your joints and muscles to move in their full range of motion is called

Studentka2010 [4]

Answer:

Dynamic flexibility

Explanation:

Dynamic flexibility can be generally defined as the ability of the body muscles and joints to move in full range of motion. High flexibility in these joints and muscles leads to the decreasing pain and injury in different parts of the body.

Proper warm up exercises are needed to be carried out that involves both the combination of controlling movements and stretching of the body, and this directly enhances the dynamic flexibility of the body.

The athletes and sports persons possesses a good dynamic flexibility of their body as they carry our different types of body exercises.

6 0

3 years ago

The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla

kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

$v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\$

To find the time we must use another kinematic equation.

$v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]$

7 0

3 years ago