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labwork [276]
3 years ago
14

A crane lifts 1800kg mass through a vertical height of 6cm in 9 seconds, Taking (g) as 10N/kg. what is the cranes power output?​

Physics
1 answer:
valkas [14]3 years ago
6 0

Explanation:

power output=(1800×10×0.06)/9=120watts

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A car starts from rest with an acceleration of 5 ft/s. What is its velocity after it has gone 600 ft?
Soloha48 [4]

Answer:

First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters

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2 years ago
____________ are used to calculate the distance a continent has moved in a year. a. Space satellites c. Laser beams b. Mirrors d
Law Incorporation [45]
<span> Space satellites, laser beams, mirrors</span> are used to calculate the distance a continent has moved in a year.

Therefore, your correct answer would be "all of the above".
4 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

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3 years ago
What type of telescope is shown in Figure 24-2
lesantik [10]
Refractor, It's a refractor-esque telescope
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Explanation:

this is my answer this is helpful for you

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