All categories

2 years ago

Calcula la resistencia de un conductor de cobre de 2 m de longitud y 0,1 m2 de sección. Resistividad del Cu = 1,7 · 10–8 Ω · m

Physics

1 answer:

Fudgin [204]2 years ago

5 0

Answer:

an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).

You might be interested in

Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.55 μC charge and flies due west at a speed of

Illusion [34]

Answer:

3.014 x 10⁻⁸ N

Explanation:

q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C

v = speed of the jet = 685 m/s

B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T

θ = angle between the magnetic field and direction of motion = 90

magnitude of the magnetic force is given as

F = q v B Sinθ

F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90

F = 3.014 x 10⁻⁸ N

6 0

3 years ago

At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)

Mama L [17]

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, $a=0.54\ m/s^2$

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

$t=\dfrac{v-u}{a}$

$t=\dfrac{28.88-25.27}{0.54}$

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

5 0

3 years ago

WILL GIVE BRAINLY!! PLEASE HELPP

LekaFEV [45]

The purpose of the machine is to leverage its mechanical advantage such that the force it outputs to move the heavy object is greater than the force required for you to input.

But there's no such thing as a free lunch! When you apply the conservation of energy, the work the machine does on the object will always be equal to (in an ideal machine) or less than the work you input to the machine.

This means that you will apply a lesser force for a longer distance so that the machine can supply a greater force on the object to push it a smaller distance. That is the trade-off of using the machine: it enables you to use a smaller force but at the cost of having to apply that smaller force for a greater distance.

The answer is: The work input required will equal the work output.

8 0

3 years ago

What is a analogy for transition metals

Ganezh [65]

A simple metallic band model is proposed for the transition metal mono antimonides, by analogy to the transition metals.

6 0

3 years ago

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

The proton mass is equal to the antiproton mass, so:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$

$K_{ap} = 80.75 MeV$

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0

3 years ago