Why do electrons move from the negative end of the tube to the positive end

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

Anna71 [15]

Answer:

The mass flow rate is 2.37*10^-4kg/s

The exit velocity is 34.3m/s

The total flow of energy is 0.583 KJ/KgThe rate at which energy leave the cooker is 0.638KW

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7 0

2 years ago

What is the distance fallen for a freely falling object 1 s after being dropped from a rest position? What is the distance for a

kondor19780726 [428]

<h2>Distance traveled in 1 second after drop is 4.9 m</h2><h2>Distance traveled in 4 seconds after drop is 78.4 m</h2>

Explanation:

We have s = ut + 0.5at²

For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²

Substituting

s = 0 x t + 0.5 x 9.8 x t²

s = 4.9t²

We need to find distance traveled in 1 s and 4 s

Distance traveled in 1 second

s = 4.9 x 1² = 4.9 m

Distance traveled in 4 seconds

s = 4.9 x 4² = 78.4 m

Distance traveled in 1 second after drop = 4.9 m

Distance traveled in 4 seconds after drop = 78.4 m

5 0

3 years ago

The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g.

Vadim26 [7]

Answer:

7.9 $\frac{gr}{cm^3}$

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7 $cm^3$ ).

Density is $\frac{mass}{volume}$ .

So the density of that piece of metal is $\frac{55.3g}{7cm^3}$

Which leaves us with a final density of 7.9 $\frac{gr}{cm^3}$

6 0

3 years ago

Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?

jasenka [17]

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

8 0

2 years ago

A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time

saul85 [17]

Answer:

t = 0.714 s and x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

vₓ = 7.0 m / s

Let's find the time it takes to get to the river

y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{\frac{2y_o}{g} }$

t = ra 2 2.5 / 9.8

t = 0.714 s

the distance traveled is

x = vₓ t

x = 7.0 0.714

x = 5.0 m

3 0

3 years ago