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bazaltina [42]
3 years ago
15

A water wave travels across a 42-meter wide pond in 7.0 seconds. The speed of the wave is _____. 292 m/s 49 m/s 6.0 m/s 0.17 m/s

Physics
2 answers:
anzhelika [568]3 years ago
8 0

Answer:

6.0 m/s

Explanation:

A water wave travels across a 42-meter wide pond in 7.0 seconds. The speed of the wave is _____.

292 m/s

49 m/s

6.0 m/s

0.17 m/s

tekilochka [14]3 years ago
7 0
           Speed  =  (distance covered) / (time to cover the distance).

The speed of anything that covers 42 meters in 7 seconds is

               (42 meters)  /  (7.0 seconds)

           =    (42 / 7.0)      (meters/second)

           =             6.0 m/s . 
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Mechanical advantage of a simple machine is 4,what does it mean?​
adelina 88 [10]

Answer:

I hope this answer is correct

Explanation:

The mechanical advantage of a machine is 4. Mechanical advantage MA is the ratio of output (generated by the machine) force to input (applied to the machine) force. So MA = 4 means that for example if you apply 100 N then your machine will multiply that force and generate 400 N.

7 0
3 years ago
You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el
nordsb [41]

Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

  • If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

       F_{g} = F_{c} (1)

  • where Fg is the gravitational attraction, that can be written as follows        according Newton's Universal Law of Gravitation:

       F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)

  • Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming  we can treat both spheres as point charges), as follows:

       F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)

  • since m₁ = m₂ = 0.0024 kg, and  r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

       G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)

  • Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

       Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)

  • Since both charges are the same, the charge on each sphere is just the square root of (5):
  • Q = 2.066* 10⁻¹³ C.

2)

  • Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
  • Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
  • n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)
4 0
3 years ago
During nuclear fission, great amounts of energy are produced from
stiks02 [169]
D.very small amounts of mass.
6 0
3 years ago
Read 2 more answers
17 points
const2013 [10]

The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>
  • MV = mv................. Equation 1

<h3>Where:</h3>
  • M = mass of the cannon
  • m = mass of the cannonball
  • V = velocity of the cannon
  • v = velocity of the cannonball

Make v the subject of the equation.

  • v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>
  • M = 500 kg
  • V = 3 m/s
  • m = 10 kg.

Substitute these values into equation 2.

  • v = (500×3)/10
  • v = 150 m/s.

 

Hence, The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

3 0
2 years ago
When this circuit is closed, which way do the electrons flow?
gizmo_the_mogwai [7]

Answer:

from the positive end of the battery through the capacitor through the resistors to the negative end...

Current flows from higher potential (+) to lower potential (-)..

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