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AnnZ [28]
3 years ago
13

An ice skater weighs 500 [N]. He is coasting to the right at a constant velocity of 2 [m/s]. Assume

Physics
1 answer:
luda_lava [24]3 years ago
4 0

Answer:

The net force on the skater is zero. (F_{net} = 0\,N)

Explanation:

According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. (F_{net} = 0\,N)

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What minimum speed must the rocket have just before impact in order to save the explorer's life?
34kurt
<span>Answer: Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by h = 0.5*g*T^2 However, the speed of the package-rocket system must be sufficient to cross the river in that time v2 = L/T Conservation of momentum says that m1*v1 = (m1 + m2)*v2 where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system. Expressing v2 in terms of v1 v2 = m1*v1/(m1 + m2) and then expressing the time in terms of v1 T = (m1 + m2)*L/(m1*v1) substituting T in the first expression h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2 solving for v1, the speed before impact is given by v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1</span>
8 0
3 years ago
Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2
wolverine [178]

The particles can undergo small oscillations around x₂.

The given parameters;

  • <em>initial energy of the particles = E₁</em>
  • <em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

  • The maximum position the particle can oscillate is x₅
  • The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

3 0
2 years ago
A negative test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy.
dalvyx [7]

Answer: higher and lower

Explanation:

 charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential the direction of decreasing electric potential energy is the direction of increasing electric potential.

5 0
3 years ago
On isolated ground receptacles, the metal yoke ____ integrally bonded to the equipment grounding terminal of the receptacle.
Ede4ka [16]

On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.

Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.

Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.

The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.

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4 0
1 year ago
Electricity is distributed from electrical substations to neighborhoods at 13000 V. This is a 60 Hz oscillating (AC) voltage. Ne
Levart [38]

Answer:

the number of turns in the primary coil is 13000

Explanation:

Given the data in the question;

V₁ = 13000 V

V₂ = 120 V

N₁ = ?

N₂ = 120 turns

the relation between the voltages and the number of turns in the primary and secondary coils can be expressed as;

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V₁N₂ = V₂N₁

N₁ = V₁N₂ / V₂

so we substitute

N₁ = (13000 V × 120 turns) / 120 V

N₁ = 1560000 V-turns / 120 V

N₁ = 13000 turns

Therefore, the number of turns in the primary coil is 13000

8 0
2 years ago
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