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worty [1.4K]
3 years ago
9

Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in 8 s. Scenario C: 200 J of work is done in 10 s. W

hich scenario uses the most power?
Physics
2 answers:
hodyreva [135]3 years ago
8 0
Using the equation P = W/t to solve your problem . 

Thus the answer is all of them use the same amount of power. 20 J.  
Mice21 [21]3 years ago
6 0

Answer:

They all used the same power.

Explanation:

Power is equal to the ratio between the work done (W) and the time taken (t):

P=\frac{W}{t}

Let's calculate the power used in the three scenarios:

A) P=\frac{120 J}{6 s}=20 W

B) P=\frac{160 J}{8 s}=20 W

C) tex]P=\frac{200 J}{10 s}=20 W[/tex]

So, the power used is the same in the three scenarios.

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TiliK225 [7]

Answer:

a) By v^2 = u^2 + 2as => a= 70291.70.

(b)By v = u + at => t= 1.58 ms.

(c)By v^2 = u^2 - 2gh => H = 46045.92 m.

Explanation:

a) By v^2 = u^2 + 2as

(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70

(b)By v = u + at

950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms

(c)By v^2 = u^2 - 2gh

0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m

4 0
2 years ago
The variation in the pressure of helium gas, measured from its equilibrium value, is given by ΔP = 2.9 × 10−5 cos (6.20x − 3 000
nadya68 [22]

Answer:

The wavelength of this wave is 1.01 meters.

Explanation:

The variation in the pressure of helium gas, measured from its equilibrium value, is given by :

\Delta P=2.9\times 10^{-5}\ cos(6.2x-3000t)..............(1)

The general equation is given by :

\Delat P=P_o\ cos(kx-\omega t)...........(2)

On comparing equation (1) and (2) :

k=6.2

Since, k=\dfrac{2\pi}{\lambda}

\dfrac{2\pi}{\lambda}=6.2

\lambda=1.01\ m

So, the wavelength of this wave is 1.01 meters. Hence, this is the required solution.

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3 years ago
We use coal, crude oil, and natural gas to generate electricity in _____________.
tresset_1 [31]

Answer:

<u>thermal power stations</u>

Explanation:

these resources are burned to produce the electricity.

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2 years ago
Read 2 more answers
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
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3 years ago
A cat runs and jumps from one roof top to another which is 5 m away and 3 m below. Calculate the minimum horizontal speed with w
icang [17]
ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s
6 0
3 years ago
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