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3 years ago

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The mechanical energy of a bicycle at the top of a hill is 6,000 J. The bicycle stops at the bottom of the hill by applying the

brakes. If the potential energy of the bicycle is 2,000 J at the bottom of the hill, how much thermal energy was produced?

1 answer:

Semmy [17]3 years ago

7 0

Answer:

Thermal energy produce =4,000 J.

Explanation:

Given that

Mechanical energy at the top of hill = 6,000 J

Mechanical energy at the bottom of hill = 2,000 J

We know that energy is conserve

Energy at top of hill = energy at bottom of hill + Thermal energy produce

So now by putting the values

Energy at top of hill = energy at bottom of hill + Thermal energy produce

6,000 = 2,000+ Thermal energy produce

Thermal energy produce =6,000-2,000 J

Thermal energy produce =4,000 J.

As we know that thermal energy produce due to friction when one mechanical component slides on the other mechanical component then always heat is generated and this heat is known as thermal energy.

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A plane wall of thickness 0.1 mm and thermal conductivity 25 W/m K having uniform volumetric heat generation of 0.3 MW/m3 is ins

juin [17]

Answer:

The maximum temperature is 90.06° C

Explanation:

Given that

t= 0.1 mm

Heat generation

$q_g=0.3\ MW/m^3$

Heat transfer coefficient

$h=500\ W/m^2K$

Here one side(left side) of the wall is insulated so the all heat will goes in to right side .

The maximum temperature will at the left side.

Lets take maximum temperature is T

Total heat flux ,q

$q=q_g\times t$

$q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2$

$q=30\ W/m^2$

So the total thermal resistance per unit area

$R=\dfrac{t}{K}+\dfrac{1}{h}$

$R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}$

R=0.002 K/W

We know that

q=ΔT/R

30=(T-90)/0.002

T=90.06° C

The maximum temperature is 90.06° C

3 0

2 years ago

1. When you have different masses for each sphere, how does the force that the larger mass sphere exerts on the smaller mass sph

aleksandrvk [35]

1) The forces are equal (Newton's third law of motion)

2) The force between the spheres will quadruple

3) The force of gravity exerted by the notebook on you is negligible

Explanation:

1)

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

<em>"When an object A exerts a force (called </em><em>action</em><em>) on an object B, then object B exerts an equal and opposite force (called </em><em>reaction</em><em>) on object A"</em>

In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

2)

The magnitude of the gravitational force between the two spheres is given by

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

$m_1, m_2$ are the masses of the two spheres

r is the separation between the two spheres

In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

$r'=\frac{r}{2}$

Substituting into the equation, we find

$F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F$

So, the force between the two spheres will quadruple.

3)

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

$F=G\frac{m_1 m_2}{r^2}$

Where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

Let's estimate the following:

$m_1 = 60 kg$ is your mass

$m_2 = 2 kg$ is the mass of the notebook

$r=1 m$ , assuming the notebook is at 1 metre from you

Substituting,

$F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N$

We see that this force has an extremely small value: therefore, it is almost negligible in daily life, where other much stronger forces act on you.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

8 0

3 years ago

Okay i'm totally stuck and nobody I know really gets it either, so i've turned to Yahoo for help :)

OlgaM077 [116]

Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:

   (weight) x (distance from the pivot) <u>on one side</u>
is equal to
   (weight) x (distance from the pivot) <u>on the other side</u>.

That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.

   (Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).

So now we come to the strange beings on the alien planet.
There are three choices right away that both work:

<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.

   (400) x (1) = (200) x (2)

<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.

   (200) x (1) = (100) x (2)

<u>#3).</u>

On one side: (300 N) in the end-seat    (300) x (2) = <u>600</u>

On the other side:
      (400 N) in the middle-seat (400) x (1) = 400
   and    (100 N) in the end-seat    (100) x (2) = 200
Total . . . . . . . . . . . . <u>600</u>

These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.

5 0

3 years ago

Read 2 more answers

Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2

aleksandr82 [10.1K]

Answer:

The curl is $0 \hat x -z^2 \hat y -4xy \hat z$

Explanation:

Given the vector function

$\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z$

We can calculate the curl using the definition

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|$

Thus for the exercise we will have

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|$

So we will get

$\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z$

Working with the partial derivatives we get the curl

$\nabla \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z$

6 0

3 years ago

A light ray passing from medium 1 to medium 2 is bent away from the perpendicular normal to the boundary surface. What happens t

SVETLANKA909090 [29]

Answer:

The answer is a for Plato users.

Explanation:

Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.

6 0

2 years ago

Read 2 more answers