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Feliz [49]
2 years ago
15

You lift a 10-kg box to a height of 1m. How much work do you do on the box when you lift it from the ground? (g= 9.81 m/s2)

Physics
1 answer:
Andreyy892 years ago
5 0

Answer:

98.1 Joule

Explanation:

Solution,

⇒Mass(m)=10kg

⇒Weight(F)=Mg=10×9.81=98.1N

⇒Distance(d)=1m

Now,

Work done=F×d

                  =98.1×1

                  =98.1J

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When the Sun is directly overhead, the intensity of sunlight reaching the ground is about 1000 W/m^{2} 2 . The Earth has a radiu
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Answer:

The total power associated with sunlight shining on the Earth is 8.04 × 10^10W

Explanation: Please see the attachment below

6 0
3 years ago
A proton moves at constant velocity in the +y direction, through a region in which there is an electric field and a magnetic fie
Ivenika [448]

Answer:

a) F_{e} - F_{m} = 0,  b) v = 666.67 m / s

Explanation:

For the proton to move y-axis  the sum of the electric and magnetic force must be zero, therefore

        F_{e} - F_{m}= 0

a) ∑ F = 0

        F_{e} - F_{m} = 0

b) we write the forces

         q E = q v B

         v = E / B

Let's calculate

      v = 300 / 0.45

      v = 666.67 m / s

4 0
3 years ago
the process of making alloys involves _____ pure metals to remove impurities. Then the pure metals are ____ with other component
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I think this is the answer for the first line(Cooling ,Heating or mixing ) and for the second line is(broken down,cooled,mixed)
5 0
3 years ago
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A car traveling in a straight line has a velocity of 6m/s at some instant. After 6.32s its velocity is 13.2m/s . What is the ave
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Average acceleration  =  (change in speed) / (time for the change) .

Average acceleration  =  (13.2 - 6) / (6.32) = 7.2 / 6.32 = about  <em>1.139... m/s²</em> .
8 0
3 years ago
A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w
Rina8888 [55]
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r 
 v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span> <span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span> <span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min). 
 (32.68 rev/s)(60 s/min)=<span> <span>1960.74 rev/min
</span></span>
5 0
3 years ago
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