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konstantin123 [22]
3 years ago
6

The density of aluminum is 2.70 g/ml. A piece of aluminum foil has a mass of 44 g. What is the volume of this piece of aluminum

foil?
A) 0.06 ml B) 1.63 ml C) 16.3 ml D) 118.8 ml
Physics
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

C) 16.3 ml

Explanation:

Density is equal to the ratio between the mass of an object and its volume:

d=\frac{m}{V}

where

m is the mass

V is the volume

In our problem, we know:

- density of aluminium: d=2.70 g/mL

- mass of the aluminium foil: m=44 g

So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

V=\frac{m}{d}=\frac{44 g}{2.70 g/mL}=16.3 mL

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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
2 years ago
Which of the following provides evidence that there must be at least two types of electrical charge, but that there is only one
polet [3.4K]

Answer:

Option D (On the...............dominate) would be the right approach.

Explanation:

The Gravitational constant (G) will be:

= 6.67\times 10^{-11}

The Coulomb's law constant (K) will be:

= 9\times 10^9

  • Throughout particular, these have been determined that among 2 substances with almost the similar form of charge, the combination of electromagnetic as well as the force does seem to be usually the following:

⇒ \frac{f_e}{f_g}\sim 10^{42}

  • By that same argument, the electrostatic force including its planet's atmosphere would have strongly influenced the effect, as well as maybe the planet's atmosphere, would have crashed, or perhaps the earth would have shifted at a much longer exposure from one another and.
  • Throughout particular, astronomical distance statutory framework that gravity seems to be predominant, whereas electrostatic forces have been generally ignored. It is quite since there are so many categories of allegations throughout the planet's atmosphere that balance out someone else's effects, there's only yet another form of momentum, because although the forces are still cumulative, as well as therefore offering to help everything hold to the universe, encouraging the universe just to rotate across the sun.

The latter three choices aren't connected to either the situation mentioned in the clarification segment elsewhere here.

5 0
3 years ago
What is capacitance?
just olya [345]

the amount of charge stored per volt

3 0
3 years ago
A satellite is in a circular orbit around an unknown planet. The satellite has a speed of 1.75 104 m/s, and the radius of the or
Arisa [49]

Answer:

v = 1.32 10² m

Explanation:

In this case we are going to use the universal gravitation equation and Newton's second law

    F = G m M / r²

    F = m a

In this case the acceleration is centripetal

    a = v² / r

The force is given by the gravitational force

    G m M / r² = m v² / r

    G  M/r =  v²

Let's calculate the mass of the planet

    M = v² r / G

    M = (1.75 10⁴)² 5.00 10⁶ / 6.67 10⁻¹¹

    M = 2.30 10²¹ kg

With this die we clear the equation to find the orbit of the second satellite

    v = √ G M / r

    v = √ (6.67 10⁻¹¹ 2.30 10²¹ / 8.75 10⁶)

    v = 1.32 10² m

8 0
3 years ago
A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8
fomenos

Answer:

v = 186.90\,\frac{m}{s}

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h

The final velocity of the system formed by the ballistic pendulum and the bullet is:

v = \sqrt{2\cdot g\cdot h}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}

v\approx 0.78\,\frac{m}{s}

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )

v = 186.90\,\frac{m}{s}

5 0
3 years ago
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