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3 years ago

6

A substance has a melting point of 800ºC and a boiling point of 1465ºC. The substance is most likely

1 answer:

VladimirAG [237]3 years ago

8 0

It is most likely table salt.

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How does light travel?

Ivenika [448]

Answer:

C

Explanation:

light can travel in a vacuum Anne the sped varies

7 0

2 years ago

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A large lightning bolt consists of a 18.2~\text{kA}18.2 kA current that moved 30.0~\text{C}30.0 C of charge. Assuming a constant

marissa [1.9K]

Answer:

$1.65\times {-3} \text { s}$

Explanation:

Current is the rate of flow of charge.

$I=\dfrac{Q}{t}$

$t = \dfrac{Q}{I}$

$t = \dfrac{30.0}{18.2\times10^3} =1.65\times {0-3} = 1.65\times {-3} \text { s}$

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2 years ago

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a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column

scoray [572]

Answer:

The answer is " $1155\ \frac{kg}{m^3}$ "

Explanation:

Please find the complete question in the attached file.

$p = p_0 + ?gh$

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

$h_1 = 11 \ cm\\\\h_2= 3 \ cm$

The pressures would be proportional to the quantity $11-3 = 8$ cm from below the surface at the interface between both the oil and the liquid.

$\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\$

$= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}$

8 0

2 years ago

A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s

Amiraneli [1.4K]

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy

Using formula of linear kinetic energy

$K.E_{l}=\dfrac{1}{2}mv^2$

$K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2$

$K.E_{l}=1.279\ J$

We need to calculate the rotational kinetic energy

$K.E_{r}=\dfrac{1}{2}\times I\omega^2$

$K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2$

$K.E_{r}=\dfrac{1}{4}\times m\times v^2$

$K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2$

$K.E_{r}=0.639\ J$

The total kinetic energy is given by

$K.E=K.E_{l}+K.E_{r}$

$K.E=1.279+0.639$

$K.E=1.918\ J$

Hence, The cylinder’s total kinetic energy is 1.918 J.

8 0

3 years ago

A jet airliner moving initially at 548 mph

sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

$v_1 =548 mph$ along the x-direction

$v_2 = 343 mph$ in a direction $67^{\circ}$ north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

$v_{1x}= 548 mph$

$v_{1y}=0$

$v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph$

$v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph$

So, the components of the resultant velocity in the two directions are

$v_{x}=548 mph+134 mph=682 mph$

$v_{y}=0 mph+315.7 mph=315.7 mph$

So the new speed of the aircraft is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph$

3 0

3 years ago