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hoa [83]
3 years ago
7

0.75 km expressed in centimeters

Physics
2 answers:
Rufina [12.5K]3 years ago
6 0
0.75 kilometer= 75000 centimeters
AnnZ [28]3 years ago
5 0
There is 100,000 centimeters in one kilometer. Since .75 kilometers is 3/4 of one kilometer, you need to find 3/4 of 100,000. 100,000*3/4=75,000. The answer is that there is 75,000 centimeters in one kilometer. Hope this helps! ;)
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The nuclear power used for electricity is produced by
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Answer:

<h3>b.fission. </h3>

Explanation:

<h3>Please mark my answer as a brainliest.Please follow me ❤❤❤</h3>
5 0
3 years ago
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Most asteroids lie between the orbits of
Anika [276]
I believe the answer is B. that's where the asetroid belt is.
4 0
3 years ago
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In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br
Levart [38]

Answer:

(A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

\lambda=\dfrac{dy}{mD}

m=\dfrac{d y}{\lambda D}

Put the value into the formula

m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}

m=5.65 = 6

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

\beta=\dfrac{yd}{D}

Put the value in to the formula

\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}

\beta=3.33\times10^{-6}\ m

\beta=3.33\ \mu m

Hence, (A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0
3 years ago
A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
  • Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

8 0
2 years ago
Can someone please give me the (Answers) to this? ... please ...<br><br> I need help….
IceJOKER [234]

#1.

<em>Car </em>1<em> weighs </em>300 kilograms<em> and is moving right at </em>3 meters per second (m/s)

  • v1 (before) = 3 m/s

  • v2 (before) = -1 m/s

  • v1 (after) = 0.5 m/s

#2.

Law of conservation of momentum

momentum before collorion = momentim after collosion

MV + mv = MV' + mv'

1500x25+ 1000x5

37500 + 15000

6 0
2 years ago
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