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2 years ago

Is energy released or absorbed during the formation of a solution?

1 answer:

3 0

Energy can be released and absorbed during the formation of a solution, not one or the other. When a solute interacts with the solvent, energy is absorbed so the solvent can overcome the intermolecular bonds of the solute and energy is released, most commonly, in the form of heat, light, or a gaseous byproduct.

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A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram

Sauron [17]

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

$\rm F=qvBsin\theta$

where

The magnitude of the charge $q=5.7\ \mu C =5.7\times 10^{-6} \ C$

The velocity of the charge $v=4.5\times 10^5\ \frac{m}{s}$

The magnitude of the magnetic field $3.2\ mT=0.0032\ T$

The angle between the directions of v and B $\theta =90^o-37^o=53^o$

By substituting the values we will get:

$F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)$

$F=6.6\times 10^{-3}\ N$

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

To know more about Magnetic force follow

brainly.com/question/14411049

3 0

2 years ago

A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet

densk [106]

Answer:

0.8726 $m^3/s$

Explanation:

We are to convert 1.85 x $10^3$ $ft^3/min$ to $m^3/s$

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x $10^3$ ft3 = 1.85 x $10^3$ x 0.0283 m3

= 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x $10^3$ $ft^3/min$ = 52.355/60 $m^3/s$

= 0.8726 $m^3/s$

7 0

3 years ago

Riders in a carnival ride stand with their backs against the wall of a circular room of diameter

Veseljchak [2.6K]

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

= $45 \times \dfrac{2\pi}{60}$

=4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

= 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

μ = 0.11

hence, the correct answer is option C

6 0

3 years ago

A disk-shaped merry-go-round of radius 2.63 m and mass 152 kg rotates freely with an angular speed of 0.526 rev/s. A 51.7 kg per

zalisa [80]

Explanation:

Given

radius $r=2.63 m$

$N=0.526 rev/s$

$\omega =3.30 rad/s$

mass disc $M=152 kg$

mass of person $m=51.7 kg$

velocity of Person $v=2.76 m/s$

moment of inertia $I=Mr^2$

$I=0.5\times 152\times 2.63^2=827.64 kg-m^2$

Initial angular momentum

$L_i=I\omega +mvr$

$L_i=827.64\times 3.30+51.7\times 2.76\times 2.63$

$L_i=2731.212+375.27=3106.48 Js$

Final Moment inertia

$I_f=0.5Mr^2+mr^2$

$I_f=(152\cdot 0.5+51.7)\cdot (2.63)^2=1185.243 kg-m^2$

final angular momentum

$L_f=I_f\omega _f$

Conserving angular momentum

$L_i=L_f$

$3106.48=1408.97\times \omega _f$

$\omega _f=2.62 rad/s$

4 0

3 years ago

Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

<em>"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.</em>"

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "</em>

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "</em>

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago