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andreev551 [17]
3 years ago
14

You can carry a 50 N weight up a flight of stairs that is 3 m high in 10 s. Then you carry 40 N up two flights of stairs in 20 s

. Finally, you carry 30 N up three flights in 45 s. Is the job that took the most power the same as the job that took the most energy? Explain.
Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0
You can do 50 and 10 and carry the 30 and it would be in the same power so you’ll have the same energy
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A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in
Troyanec [42]

Answer:

The value is  \epsilon =  3.84 *10^{-5} \  V

Explanation:

From the question we are told that

  The diameter of the ring is  d =  18 \ mm  =  0.018 \  m

   The length of the solenoid is l = 25 \ cm  =  0.25 \ m

   The diameter of the solenoid is  D = 5.0 \ cm  = 0.05 \ m

    The number of turns is  N = 1500

   The change in  current in the solenoid is   \Delta  I   = 20 \ A

   The time taken is  \Delta  t  = 1 \ s

Generally the radius of the ring is  

     r = \frac{d}{2}

=>  r = \frac{0.018 }{2}

=>  r = 0.009 \ m

Generally the area of the ring is mathematically represented as  

      A = \pi r^2

=>   A = 3.142 *  0.009^2    

=>   A = 2.545 *10^{-4}\ m^2

Generally the induced emf is mathematically represented as

       \epsilon  =  A * \frac{dB}{dt}

Here    

         \frac{dB }{dt} =  \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}

Here  \mu_o is the permeability of free space with value  

         \mu_o =  4\pi *10^{-7} \ N/A^2

So  

     \frac{dB }{dt} =   4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}

=>  \frac{dB }{dt} =   0.150816\  T/s

So

     \epsilon =   0.150816 *  2.545 *10^{-4}

=>   \epsilon =  3.84 *10^{-5} \  V

3 0
3 years ago
A loader sack of total mass
vampirchik [111]

Question: A loader sack of total mass

is l000 grams falls down from

the floor of a lorry 200 cm high

Calculate the workdone by the

gravity of the load.​

Answer:

19.6 Joules

Explanation:

Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

W = (1×2×9.8)

W = 19.6 Joules

Hence the work done by gravity on the load is 19.6 Joules

8 0
3 years ago
a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w
nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load  distance=10 cm

We have to find the effort distance.

We know that

load\times load\;distance=Effort\times effort\;distance

Using the formula

800\times 10=200\times effort\;distance

Effort distance=\frac{800\times 10}{200}

Effort distance=\frac{8000}{200}

Effort distance=40 cm

Hence,  the effort distance will be 40 cm.

7 0
3 years ago
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2) they add the insulation to better the durability or the earbuds

6 0
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s344n2d4d5 [400]

Answer:

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Explanation:

8 0
3 years ago
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