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3 years ago

Which parts of the graph would increase if all of earth's polar ice melts? Which part would decrease?

Physics

1 answer:

Sedbober [7]3 years ago

8 0

Answer:

the answer to the question is the part that is reflected by clouds and atmosphere 26%

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A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?

poizon [28]

Answer:

$\boxed {\boxed {\sf 7.84 \ Joules}}$

Explanation:

The formula for potential energy is:

$PE=m*g*h$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

$m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m$

Substitute the values into the formula.

$PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)$

Multiply the first two numbers.

0.4 kg*9.8 m/s²= 3.92 kg*m/s²
If we convert the units now, the problem will be much easier later on.
1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

$PE=(3.92 \ N )(2 \ m)$

Multiply.

3.92 N* 2 m=7.84 N*m
1 Newton meter is equal to 1 Joule (this is why we converted the units).
Our answer is equal to 7.84 Joules.

$PE=7.84 \ J$

6 0

3 years ago

Read 2 more answers

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

2. A ball is released from a vertical height of 20 cm. It rolls down a "perfectly

Paraphin [41]

Answer:

20 cm

Explanation:

Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?

Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.

Therefore, the ball will reach 20 cm before it starts to roll back down.

7 0

3 years ago

Officials in the House of Representatives serve __ terms.

Salsk061 [2.6K]

Officials in the House of Representives have are able to serve a 2 year term and there are no limits as to how many terms.

7 0

3 years ago

Read 2 more answers