What causes global winds?

If there is a negative sign in front of the Hooke's Law equation, what does the force, F, represent?

SashulF [63]

Answer:

Hooke's law, given by $F=-kx$ is denoted with a negative sign to regard that the direction of restoring force that occurs opposite in direction of the force that caused the displacement $x$ . F represents the force required to create a spring displacement of $x$ .

Explanation:

As with many things in physics, the negative sign represents direction. In this case, after a spring is displaced a distance of $x$ from equilibrium, the restoring force created by the spring will be in opposite direction of the force applied to move the spring there, represented by F.

6 0

3 years ago

Describe the electron's path in this field. Describe the electron's path in this field. The electron will move in a clockwise ci

Jlenok [28]

Answer: D - The electron will move in a counterclockwise circular path if viewed in the direction of the magnetic field.

Explanation:

When the electron enters the field at an angle to the field direction the resulting path of the electron will be helical. Such motion occurs above the poles of the Earth where charges particles from the Sun spiral through the Earth's field to produce the aurorae.

Magnetic force on an electron = BILsinØ = B[e/t][vt] = BevØ

where v is the electron velocity

In a magnetic field the force is always at right angles to the motion of the electron (Fleming's left hand rule) and so the resulting path of the electron is circular.

4 0

3 years ago

lightning and thunder both occur at the same time why do you see the Lightning before you hear the thunder

denis23 [38]

Because the light from it travels to you about 874 thousand times
as fast as the sound does, so the hearing part falls behind the seeing
part.

5 0

3 years ago

Read 2 more answers

The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.

oksano4ka [1.4K]

Answer:

$i = 2.84 \times 10^{-3} A$

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

$J = \frac{di}{dA}$

so the current through the wire is given as

$i = \int J dA$

now we have

$i = \int_{0.921R}^R J dA$

here we have

$J = (3.25 \times 10^8)r^2$

now plug in the values in above equation

$i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr$

now we have

$i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr$

$i = (2.04 \times 10^9) \frac{r^4}{4}$

now plug in both limits as mentioned

$i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})$

$i = (2.04\times 10^9)(0.07 R^4)$

here R = 2.11 mm

$i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)$

$i = 2.84 \times 10^{-3} A$

8 0

3 years ago

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.60 m/s. How

kirill115 [55]

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

$E_{pot}-E_{friction}=E_{kin}$

where:

$E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}$

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

$(56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ]$

4 0

3 years ago