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liraira [26]
3 years ago
13

What were the models of ancient greek atoms ?

Physics
1 answer:
Kazeer [188]3 years ago
8 0
Democritus was the ancient greek atoms.
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One of the checks that you could do for problem 1) would be to check the output resistance of the Wheatstone bridge to make sure
Ray Of Light [21]

Answer:

Explanation:

Let the four resistances of th wheat stone bridge is

P, Q, R and S and the value of each is 350 ohm.

Here, P and Q are in series.

R' = P + Q = 350 + 350 = 700 ohm

Then R and S are in series

R' = R + S = 350 + 350 = 700 ohm

Now R' and R'' are in parallel.

So, the equivalent resistance is

Req = R' x R'' / ( R' + R'')

Req = 700 / 2 = 350 ohm

Thus, the reading of ohmmeter is 350 ohm.

6 0
3 years ago
A diver is is pushed upwards by a diving board. She weighs 739 N and accelerates from rest to a speed of 4.60 m/s while moving 0
gavmur [86]

Answer:

Answer A

Explanation:

It seems more likely

8 0
3 years ago
A current of 2 A flows through a resistor. The voltage across the resistor is 18 V.
stepan [7]

Answer:

R=9\ \Omega

Explanation:

Given that,

Current, I = 2 A

Voltage across the resistor, V = 18 V

We need to find the value of resistance of the resistor. Let the resistance be R. We can find it using Ohm's law i.e.

V = IR

Where

R is the resistance of the resistor

R=\dfrac{V}{I}\\\\R=\dfrac{18}{2}\\\\R=9\ \Omega

So, the resistance of the resistor is equal to 9\ \Omega.

3 0
3 years ago
The mass of an object is 275.32g and its density is 7.562g/cm 3 . Calculate its volume by keeping significant figures in view.
TEA [102]

Answer:

36.408cm3

Explanation:

Since we acknowledge that density is d= m/v, once we switch it up to maintain v as the number to be found it will change to v=m/d. Therefore, 275.32/7.562 is 36.408 and the unit is cm cube!

Hope that helped!!

5 0
3 years ago
A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high
Lelu [443]

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}

Where:

v_{o} - Initial velocity, measured in meters per second.

y - Final height, measured in meters.

y_{o} - Initial height, measured in meters.

t - Time, measured in seconds.

g - Gravitational constant, measured in meters per square second.

(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}

Given that y = 0\,m, v_{o} = 15\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} and t = 16\,s, the final height of the package is:

y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}

y_{o} = 1015.296\,m

The package was released at a height of 1015.296 meters.

7 0
3 years ago
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