What is the frequency of a photon with an energy of 1.99 x 10-19 j?

Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur

Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

$I_1$ is the current in the 1st wire

$I_2$ is the current in the 2nd wire

r is the separation between the wires

In this problem

$I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m$

Substituting, we find the force per unit length on the two wires:

$\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N$

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

3 0

3 years ago

The joule and the kilowatt-hour are both units of energy. 15 kw · h is equivalent to how many joules? answer in units of j.

choli [55]

The solution for the problem is:

1 Watt = 1 Joule per second
1 Watt*second = 1 Joule

a Kilowatt is 1,000 Watts
an hour is 60 seconds times 60 minutes or 3,600 seconds
a Kilowatt * hour is 1,000 Watts in 3,600 seconds

15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000 Watt*second

54,000,000 Watt*second = ? Joules
54,000,000 Joules / second = 54,000,000 Watts

3 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago

A disk of radius R = 9.54 cm, is centered at the origin and lies along the yâ€“z plane. The disk has a surface charge density Ïƒ

lyudmila [28]

Answer:

Electric field by charged disk is given as

E = (Charge Density/2u0)*[1 - (z/sqrt(z^2 - R^2))]

R = 9.54cm = 0.0954m, z = 1.01m, Charge density = 4.07 x 10^-6C/m2, e0 = 8.85 x 10^-12F/m.

Substituting all the values in to equation,

E = (2.299 x 10^5) x (8.931 x 10^-3)

E = 2.053 x 10^3N/C

Explanation:

3 0

3 years ago

A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou

prisoha [69]

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

$tan\theta =\frac{v^{2}}{rg}$

$tan\theta =\frac{18.61^{2}}{53.1\times 9.8}$

tan θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

6 0

3 years ago