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Vilka [71]
3 years ago
12

An object has a mass of 15 kg. on planet x, the object weighs 126 n. the force of gravity on planet x is _____. 0.12 m/s2 8.4 m/

s2 1875 m/s2
Physics
2 answers:
alexandr402 [8]3 years ago
4 0
<span>8.4 m/s^<span>2
Is correct!</span></span>
Lana71 [14]3 years ago
4 0

Answer:

B. 8.4 m/s^2

Explanation:

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Which of these warnings refers to a chemical property of the material?
Wewaii [24]

Answer:

Please show the warning's

Explanation:

6 0
3 years ago
4. What is the momentum of a 70 kg object traveling at 20 m/s?
Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

6 0
3 years ago
Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus
poizon [28]

Answer:

For destructive interference phase difference is

(2n+1)\pi where n∈ Whole numbers

Explanation:

For sinusoidal wave the interference affects the resultant intensity of the waves.

In the given example we have two waves interfering at a phase difference of \frac{\pi}{4} would lead to a constructive interference giving maximum amplitude at at the RMS value of the amplitude in resultant.

Also the effect is same as having a phase difference of  ( \frac{\pi}{4} + 2\pi) because after each 2π the waves repeat itself.

<em>In case of destructive interference the waves will be out of phase i.e. the amplitude vectors will be equally opposite in the direction at the same place on the same time as shown in figure.</em>

They have a phase difference of \pi or which is same as (2\pi+\pi)

Generalizing to:

a phase difference of (2n+1)\pi where n∈ {W}

{W}= set of whole numbers.

3 0
3 years ago
A 6.00 A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5×1028 free elect
navik [9.2K]

Answer:

Explanation:

Current, I = 6 A

diameter of wire, d = 2.05 mm

number of electrons per unit volume, n = 8.5 x 10^28

If the diameter is doubled,

The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is  one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.  

8 0
3 years ago
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)
7 0
3 years ago
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