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3 years ago

A. 2kg B. 6kg C. 8kg D. 14kg

Physics

1 answer:

Sati [7]3 years ago

7 0

Answer:

Explanation:

We know the force is (16N-12), which is 4N, and we know the acceleration is 2 m/s^2. Meaning, we can solve the formula m = F / a (mass equals force divided by acceleration), and we get 2kg.

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Which tools would sara us to find an irregularly shaped objects mass and volume A. a meter stick , a ruler , and water

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I think it's D. Usually, to find the volume of an irregularly shaped object, you put it in water with a labeled beaker to measure how much the water rises. The balance would be used to measure the mass in grams.

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2 years ago

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jonny [76]

Answer:

about half the people who develop the condition survive.

Explanation:

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3 years ago

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jeka94

Answer:

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3 years ago

A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

Fantom [35]

Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

revolutions per second = 1.80 × $10^{6}$

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

- q × E = m × w² × r ..................1

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}r}$ q = m × w² × r ............2

and w = $\frac{2*\pi}{T}$

w = $\frac{d\theta }{dt}$

w = 1.80 × $10^{6}$ × $\frac{2*\pi}{1}$

w = 11304000 rad/s

so here from equation 2

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}{0.012}$ 1.80 × $10^{6}$ = 1.672 × $10^{-27}$ × 11304000² × 0.0120

linear charge density = -9.495 × $10^{-34}$ C/m

8 0

3 years ago

A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea

Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$

4 0

1 year ago