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3 years ago

What provides and has always provided most of Earth's energy for living things and technology?

1 answer:

6 0

Hints:
The Earth's energy does not come from eating a balanced diet,
or from ninja turtles, or from getting sleep and exercise.
There are no hamsters running on a treadmill inside the Earth.
God didn't wind up a big spring to operate the Earth.

The Sun provides, and has always provided, almost all of the energy
used on Earth. It was the energy that got stored in dead dinosaurs
to make gas and oil, and it's the energy that makes plants grow, and
gets stored in them so we can get the energy by eating the plants.

You might be interested in

What must you know to find the amount of work done on an object

Cerrena [4.2K]

Answer:

The work is calculated by multiplying the force by the amount of movement of an object (W = F * d). A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work. A newton-meter is the same thing as a joule, so the units for work are the same as those for energy – joules.

Explanation:

5 0

3 years ago

Read 2 more answers

A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th

Reika [66]

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

The magnitude of the resultant of two vectors is given by

$R=\sqrt{A^2 +B^2 +2AB cos \theta}$

where

A is the magnitude of the first vector

B is the magnitude of the second vector

$\theta$ is the angle between the directions of the two vectors

In the formula, A and B are constant, so the behaviour depends only on the function $cos \theta$ . The value of $cos \theta$ are:

- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

$R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B$

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

$R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}$

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

$R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B$

4 0

3 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$