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2 years ago

What is one of the earliest ways that ancient peoples produced light?

Physics

1 answer:

VARVARA [1.3K]2 years ago

4 0

Answer:

Candles

Explanation:

The answer is the first option or "candles." In ancient times the people would lit candles and place them in difference places in their homes, temples, etc... to produce light. It's not option B because light emitting diodes are used to light up toys and such which weren't invented in ancient times. It's not option C because fluorescent bulbs are long light bulbs used to light up huge spaces without as many light bulbs which were again not invented at the time, and it's also not option D because incandescent bulbs are you usual light bulbs which weren't invented at the time.

Hope this helps.

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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)

SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c² => c = 0.707 m

Charge to the right: In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x $10^{9}$ N m²/C²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below: For y direction

e = kq/r²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0

3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t

borishaifa [10]

Answer:

Angular acceleration will be $18.84rad/sec^2$

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity $\omega _i=0rev/sec$

And final angular velocity $\omega _f=24rev/sec$

Time is given as t = 8 sec

From equation of motion

We know that $\omega _f=\omega _i+\alpha t$

$24=0+\alpha \times 8$

$\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2$

So angular acceleration will be $18.84rad/sec^2$

4 0

3 years ago

Why do the magnetic forces acting on the coil change as the current running through the coil changes?

VladimirAG [237]

Answer: The changing magnetic field caused by the material's motion induces a current in the coil of wire proportional to the change in field. If a 0 is represented, the magnetic field does not change between the two domains of a bit, so no current is induced as the magnetic material passes the coil.

4 0

3 years ago

Please help!

Yanka [14]

Answer:

V=3.57 × 10^6 m/s

hope it is helpful..

7 0

2 years ago