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Ierofanga [76]
3 years ago
15

A crane uses a block and tackle to lift a 2200N flagstone to a height of 25m

Physics
1 answer:
Cloud [144]3 years ago
6 0

Remember the headline:  ENERGY IS NEVER CREATED OR DESTROYED

The amount of energy before and after are always equal.  All we ever do with energy is move it around from one place to another.

a). A crane can't create energy.  Lifting the same rock in 20 different ways always takes the <u><em>same amount of work</em></u>.  It doesn't matter whether one person picks the rock straight up, or 50 people get around it and lift it, or roll it up a ramp, or lift it with 16 pulleys and a mile of rope, or use a giant steam crane.

You want to lift a 2200N weight up 25m, you're going to have to supply

(2200N) x (25m) = <em>55,000 Joules</em> of work.

c). YOU put out 55,000 Joules of energy.  It had to GO someplace. Where is it now ? ===>  It's the potential energy the rock has now, from being 25m higher than it was before.  That <em>55,000 Joules</em> is NOW the potential energy  of the rock.

No energy was created or destroyed.  It just got moved around.  

55,000 Joules of energy began as nuclear energy in the core of the sun. Solar radiation carried it to the Earth. Plants absorbed it, and stored it as chemical energy.  You ... or a cow that you ate later ... ate the plants and took the chemical energy.  One way or the other, the chemical energy got stored in your blood and fat.  When you needed to put it out somewhere, you moved it into your muscles, and they converted it into mechanical energy.  Then you used the mechanical energy to exert forces.  Today, you used the original 55,000 joules to lift the flagstone, and NOW that energy is in the flagstone, 25 meters up off the ground !

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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
3 years ago
What is the specific heat capacity of water?
Finger [1]
The specific heat of water is 4.186.
8 0
3 years ago
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
Which process usually requires water?A.Physical weatheringB.ErosionC.Mass wastingD.Chemical weathering
MAXImum [283]

Answer:

B erosion

Explanation:

Erosion would usually require water

8 0
3 years ago
A car advertisement states that a certain car can accelerate from rest to
kaheart [24]

Answer:

a = 2.22 [m/s^2]

Explanation:

First we have to convert from kilometers per hour to meters per second

40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]

We have to use the following kinematics equation:

v_{f}= v_{i}+(a*t)

where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

6 0
2 years ago
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