What type of motion occurs when an object spins around an axis without altering its linear position?

A wave propagates at a well defined velocity that depends on the properties of the medium that carries the wave. True or false?

likoan [24]

Answer:

True

Explanation:

A wave is generated as a result of oscillations which creates disturbances in the medium and these disturbances termed as waves propagates or travels from one point to another.

Waves can be classifies as:

Mechanical waves which requires material medium for their propagation

Electromagnetic waves which do not require any material medium to propagate.

A wave travels at a specific velocity depending on the type of the medium in which it propagates.

7 0

4 years ago

There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12

Licemer1 [7]

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

C = $\frac{Q}{\Delta V}$

C = ε₀ $\frac{A}{d}$

we solve for the charge (Q)

$\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}$

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

Q = $\epsilon_o \ \frac{A \ \Delta V_1 }{d_1}$

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

Q = \epsilon_o \ \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

$\frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}$

ΔV₂ = $\frac{d_2}{d_1 } \ \Delta V_1$

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

Em₀ = U = q DV₂

Em₀ = q \frac{d_2}{d_1 } \ \Delta V_1

final point. Proof load on the right plate

Em_f = K

energy is conserved

Em₀ = em_f

q \frac{d_2}{d_1 } \ \Delta V_1 = K

we calculate

K = 1 10⁻⁹ 12 $\frac{0.005}{0.003}$

K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

3 0

3 years ago

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

6 0

3 years ago

What type of image can be larger or smaller than the object?

Stolb23 [73]

It’s D. An enlargement (hope this helps!)

4 0

3 years ago

1.

rusak2 [61]

Answer:

9.6

Explanation:

to convert km to miles multiply by 1.609

7 0

4 years ago