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2 years ago

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what are the four things that affect the resistance of a wire? A. length, diameter, material, and temperature B. weight, diamete

r, material, and temperature C. length, height, material, temperature D. length, weight, material, and temperature

1 answer:

Airida [17]2 years ago

5 0

Answer:

A. length, diameter, material, temperature

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What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (see section 17.6;

antiseptic1488 [7]

<h3><u>Answer;</u></h3>

volume = 6.3 × 10^-2 L

<h3><u>Explanation</u>;</h3>

Volume = mass/density

Mass = 0.0565 Kg,

Density = 900 kg/m³

= 0.0565 kg/ 900 kg /m³

= 6.3 × 10^-5 M³

but; 1000 L = 1 m³

Hence, <u>volume = 6.3 × 10^-2 L</u>

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3 years ago

Which statement accurately describes a sample of water during parts a and c of the heating curve

vivado [14]

Answer:

A and C is about 12 cm away from each other.

Explanation:

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3 years ago

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Which of the following changes would make a heat engine less efficient

DIA [1.3K]

A heat engine would be less efficient due to many factors
For instance, a heat engine is more efficient when it uses in cold weather because there is a greater temperature difference ( Carnot Efficient )
A heat engine could be less efficient because of friction
Hope it helps I am a beginner

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3 years ago

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

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3 years ago

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is

Soloha48 [4]

Answer: $V=5839.051m/s$

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;:

$T=1.26(10)^{4}s$ is the period of the satellite

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=5.98(10)^{24}kg$ is the mass of the Earth

$a$ is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity $V$ is given by:

$V=\sqrt{\frac{GM}{a}}$ (2)

Now, from (1) we can find $a$ , in order to substitute this value in (2):

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$ (3)

$a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}$ (4)

$a=11705845.57m$ (5)

Substituting (5) in (2):

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}$ (6)

$V=5839.051m/s$ (7) This is the speed at which the satellite travels

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3 years ago