The state of matter that has particles in a fixed position, with a definte volume and shape is

Artemon [7]

Something that there is only so many of or not enough of

7 0

3 years ago

A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire

Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

$\theta$ is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

$\theta=90^{\circ}$ since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

$B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T$

3 0

3 years ago

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago

By using electroscope, if the body is negatively charged then due to electrostatic induction, then on the leaves there will be a

Kryger [21]

(b) negative charge

This is the answer

6 0

2 years ago

Read 2 more answers

While in empty space, an astronaut throws a ball at a velocity of 11 m/s. what will the velocity of the ball be after it has tra

Aliun [14]

In empty space probably means, there is no force on the ball.

(This assumption is not quite correct since there is still the force of gravity between the ball and the astronaut, but this force is very very small and can be neglected.)

Assuming there is no force on the ball, Newtown's 1st law says: When viewed in an internal frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.

This means:
If there is no force on the ball, there will be no acceleration on the ball either.
If the acceleration is zero, the velocity of the ball never changes.

3 0

3 years ago