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3 years ago

Which of the following correctly compares the uses of bar graphs versus pie charts?

1 answer:

6 0

Bar graphs show data involved in distinct categories that do not overlap; whereas, pie charts show data as parts out of a whole (such as out of 100%).

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a small asteroid of mass 125 kg is orbiting a planet that has a mass of 3.52x 10^13 what is the radial distance between the aste

asambeis [7]

Answer:

r = 2.031 x 10⁶ m = 2031 km

Explanation:

In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:

Centripetal Force = Gravitational Force

mv²/r = GmM/r²

v² = GM/r

r = GM/v²

where,

r = radial distance = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Planet = 3.52 x 10¹³ kg

v = tangential speed = 0.034 m/s

Therefore,

r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²

7 0

3 years ago

A capacitor has a capacitance of 0. 40 µF at a voltage of 9. 0 V. What is the charge on each plate of the capacitor? µC.

ASHA 777 [7]

The capacitor is a device that can store electrical energy. It is a two-conductor configuration. The charge on each plate of the capacitor will be 3.6 µC.

<h3>What is a capacitor?</h3>

A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.

An electric insulator or vacuum, such as glass, paper, air, or a semi-conductor termed a dielectric, can be used as the non-conductive zone.

The given data in the problem is;

C is the capicitence of capicitor= 0. 40 µF

V is the voltage = 9. 0 V

Q is a charge on each plate of the capacitor=?µC.

The formula for the capacitor is given as;

$\rm Q=CV \\\\ \rm Q=0. 40 \times 9. 0 \\\\ \rm Q=3.6 \ \mu C.$

Hence the charge on each plate of the capacitor will be 3.6 µC.

To learn more about the capacitor refer to the link;

brainly.com/question/14048432

4 0

2 years ago

Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele

Marta_Voda [28]

Answer:

b) True Only if the finger is isolated from ground

c) True. The total charge does not change since the system is isolated

Explanation:

When the electroscope is touched with an acrylic rod, some charges are transferred from the electroscope to the rod, until the charge in both is equal.

In the case it know when the electroscope is touched with a finger, two things can happen.

- The body is isolated from the ground, the efective charge is redistributed between the two bodies. Case similar to insulating rod

- The body is connected to ground, the charge is transferred to the finger and from here to the ground until the total charge is transferred and the Earth and the final charge of the electroscope is zero.

Let's review the final statements

a) False, when part of the load is touched, it passes to the rod, so when it separates it does not return to the initial load

b) True Only if the finger is isolated from ground

c) True. The total load does not change since the system is isolated

d) False. The value of the load changes =, but its sign does not

8 0

3 years ago

A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor

natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = $120\Omega$

Fixed resistance, R = $120\Omega$

Gauge Factor, G.F = 2.01

Supply Voltage, $V_{s} = 3\ V$

Strain, $\epsilon = 1000\times 10^{-6}\ strain$

Now,

To calculate the output voltage, $V_{o}$ :

WE know that strain is given by:

$\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}$

Thus

$V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}$

Now, substituting the suitable values in the above eqn:

$V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}$

$V_{o} = 1.507\ mV$

6 0

3 years ago

What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?

kipiarov [429]

Answer:

$\boxed {\boxed {\sf a= -7 \ m/s^2}}$

Explanation:

Acceleration is the change in velocity over time.

$a= \frac {v_f-v_i}{t}$

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

$v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s$

Substitute the values into the formula.

$a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}$

Solve the numerator.

$a= \frac { -35 \ m/s}{5 \ s}$

Divide.

$a= -7 \ m/s/s$

$a= -7 \ m/s^2$

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0

3 years ago