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Cloud [144]
3 years ago
11

What is the name of the North Star?

Physics
1 answer:
meriva3 years ago
5 0
<span>Polaris might be the answer you are looking for</span>
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A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an angle of 35°
spin [16.1K]

<em>Answer</em>


Force = 170 N



<em>Explanation</em>

First find the distance (d) travelled by the bulldozer.


Sin 35 = 15/d

d = 15/(sin 35)

= 26.15m


Now;

work done = force × distance.


4500 J = force × 26.15


dividing both sides by 26.15,


Force = 4500/26.15

= 172.07 N


Answer to two significant figures = 170 N

3 0
3 years ago
Read 2 more answers
An AWG No.16 line cord with type HPN insulation has an amp airy of
Lostsunrise [7]

12 amp is your answer

5 0
2 years ago
Jamie decides to drop an egg that has a mass of 345 g off the top of a 8.2 m building. How fast will it be falling right before
zlopas [31]

Answer:

13 m/s

Explanation:

Given:

Δy = 8.2 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (8.2 m)

v = 12.7 m/s

Rounded to two significant figures, the speed is 13 m/s.

3 0
2 years ago
Iron (Fe) is classified as a metal while Silicon (Si) is classified as a metalloid. Which physical property do these two element
Len [333]

Answer:d

Explanation:

6 0
3 years ago
Through what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinho
Mila [183]

Answer:

0.15 mV

Explanation:

In order to exhibit wave nature, the de Broglie wavelength of the electron must be of the same size of the diameter of the pinhole, therefore:

\lambda=0.10 \mu m = 1.0\cdot 10^{-7} m

The de Broglie wavelength of an electron is

\lambda = \frac{h}{mv}

where

h=6.63\cdot 10^{-34} Js is the Planck constant

m=9.11\cdot 10^{-31} kg is the mass of the electron

v is the electron's speed

Therefore, the electron's speed must be

v=\frac{h}{m\lambda}=\frac{6.63\cdot 10^{-34}}{(9.11\cdot 10^{-31})(1.0\cdot 10^{-7})}=7278 m/s

When accelerated through a potential difference \Delta V, the kinetic energy gained by the electron is equal to the change in electric potential energy, therefore

e\Delta V = \frac{1}{2}mv^2

where

e=1.6\cdot 10^{-19} is the magnitude of the charge of the electron

So, we can find the potential difference needed:

\Delta V=\frac{mv^2}{2e}=\frac{(9.11\cdot 10^{-31})(7278)^2}{2(1.6\cdot 10^{-19})}=1.5\cdot 10^{-4}V = 0.15 mV

5 0
3 years ago
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