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3 years ago

9

I Need Assistance With This Equation:

1 answer:

nikklg [1K]3 years ago

8 0

Check the picture below.

now, for 6 triangles, well, simply 6*6.

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What's the mass show the work?

Alja [10]

You do this one just like the other one that I just solved for you.

For this one ...

The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.

The question tells us: We have 2.0 cm³.

Each cm³ of space that the object occupies contains 2.5 gm of mass.

So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.

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3 years ago

I need help with this someone help me?

Brilliant_brown [7]

Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol

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2 years ago

What is the new current?

nikdorinn [45]

Voltage = current x resistance
since R is doubled, current must reduce by half.
So,
new current = 120/2 = 60mA

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3 years ago

A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart

Musya8 [376]

1) $0.61 m/s^2$

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

$g=\frac{GM}{r^2}$ (1)

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

$r=4R$

where

$R=6.37\cdot 10^6 m$ is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

$g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2$

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

$g=0.61 m/s^2$ is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

$T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s$

4 0

3 years ago

What material would make the best light bulb filament?

DanielleElmas [232]

The answer is B ! As in light bulb tungsten is working as resistor !

7 0

3 years ago

Read 2 more answers