How are the electric field lines around a positive charge affected when a second positive charge is near it?

Vectors A and B lie in the xy-plane. Vector A has a magnitude of 19.1 and is at an angle of 125.5º counterclockwise from the +x-

Nady [450]

Answer:

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Explanation:

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8 0

3 years ago

Which of the following is not a an example of dissipated energy?

Kaylis [27]

Which of the following is not a an example of dissipated energy?
b. kinetic

When energy is changed from one form to another, ____.

b. all of the energy can be accounted for

4 0

2 years ago

Which of the following is NOT true? a. The higher the moment of inertia, the greater the resistance to changes in angular veloci

kaheart [24]

Answer:

Explanation:

The moment of inertia is the integral of the product of the squared distance by the mass differential. Is the mass equivalent in the rotational motion

a) True. When the moment of inertia is increased, more force is needed to reach acceleration, so it is more difficult to change the angular velocity that depends proportionally on the acceleration

b) True. The moment of inertia is part of the kinetic energy, which is composed of a linear and an angular part. Therefore, when applying the energy conservation theorem, the potential energy is transformed into kinetic energy, the rotational part increases with the moment of inertia, so there is less energy left for the linear part and consequently it falls slower

c) True. The moment of inertial proportional to the angular acceleration, when the acceleration decreases as well. Therefore, a smaller force can achieve the value of acceleration and the change in angular velocity. Consequently, less force is needed is easier

4 0

2 years ago

An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligibl

krok68 [10]

(a) The moment of inertia of the wheel is 78.2 kgm².

(b) The mass (in kg) of the wheel is 1,436.2 kg.

(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

<h3>Moment of inertia of the wheel</h3>

Apply principle of conservation of angular momentum;

Fr = Iα

where;

F is applied force
r is radius of the cylinder
α is angular acceleration
I is moment of inertia

I = Fr/α

I = (200 x 0.33) / (0.844)

I = 78.2 kgm²

<h3>Mass of the wheel</h3>

I = ¹/₂MR²

where;

M is mass of the solid cylinder
R is radius of the solid cylinder
I is moment of inertia of the solid cylinder

2I = MR²

M = 2I/R²

M = (2 x 78.2) / (0.33²)

M = 1,436.2 kg

<h3>Angular speed of the wheel after 4 seconds</h3>

ω = αt

ω = 0.844 x 4

ω = 3.376 rad/s

Thus, the moment of inertia of the wheel is 78.2 kgm².

The mass (in kg) of the wheel is 1,436.2 kg.

The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

Learn more about moment of inertia here: brainly.com/question/14839816

#SPJ1

7 0

1 year ago

You are the juror of a case involving a drunken driver whose 1041 kg sports car ran into a stationary 1928 kg station wagon stop

mina [271]

1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g
2. Find the acceleration due to friction:
F = m*a = μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t²
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:

m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash

m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s

7 0

3 years ago