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3 years ago

How could a sedimentary rock provide evidence that the rock cycle exists?\?

Physics

1 answer:

worty [1.4K]3 years ago

3 0

If im right the answer is wrong actually because no clue

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From inside to out, describe the components of an atom. What makes one element's atoms differ from another element's atoms?

Vesna [10]

From the outside is it’s shell, and on the inside is the nucleus (on the middle) which is made or protons (positive charge) and neutrons (neutral charge which is negate and positive) and Lastly there are electrons (negative charge) that orbit the nucleus like planets orbiting the Sun. The shell gets made as the electrons orbit the nucleus.
How do atoms differ from eachother?- their atoms might have different elections, neutrons, and/ or protons inside of it. Plus the location of the elections might be different too.

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4 years ago

Position vs. Time

olga nikolaevna [1]

Answer:

Explanation:

cause the slope works with going up or down

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2 years ago

Mark is biking a a velocity of 20 m/s and has a mass of 70kg what is his momentum

Snowcat [4.5K]

Momentum=Mass×Velocity

20×70=1400 (Kg×M)÷S

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4 years ago

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If th

tensa zangetsu [6.8K]

Answer:

The net charge is $67.89 \mu C$

Solution:

As per the question:

Mass of the plastic bag, m = 12.0 g = $12.0\times 10^{-3}\ kg$

Magnitude of electric field, E = $10^{3}\ N/C$

Angle made by the string, $\theta = 30^{\circ}$

Now,

To calculate the net charge, Q on the ball:

Vertical component of the tension in the string, $T = Tcos\theta$

Horizontal component of the tension in the string, $T = Tsin\theta$

Now,

Balancing the forces in the x-direction:

$Tsin\theta = QE$

$Q = {Tsin\theta}{E}$ (1)

Balancing the forces in the y-direction:

$Tcos\theta = mg$

where

g = acceleration due to gravity = $9.8\ m/s^{2}$

Thus

$T = \frac{mg}{cos\theta }$

$T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N$

Use T = 0.1357 N in eqn (1):

$Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C$

$Q = 67.89\times 10^{- 5}\ C = 67.89\mu C$

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4 years ago

How much heat is contained in 100 kg of water at 60.0 °C?

attashe74 [19]

First we need to write down heat capacity for water which is constant.
cp=4186 J/(kg*K)

The equation for Energy that we will be calculating is:

E=cp*m*T
where m is mass and T is absolute temperature (273,15 + 60 in this case). Replacing all the values in equation we get:

E = 4186*100*333,15 = 139 456 590 J

6 0

3 years ago