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maxonik [38]
3 years ago
13

An electromagnet produces a magnetic field of 0.520 T in a cylindrical region of radius 2.40 cm between its poles. A straight wi

re carrying a current of 10.5 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field.
What magnitude of force is exerted on the wire?
Physics
1 answer:
koban [17]3 years ago
3 0

Answer:

Magnetic force, F = 0.262 N

Explanation:

It is given that,

Magnetic field of an electromagnet, B = 0.52 T

Length of the wire, l = 2r =2\times 2.4=4.8\ cm=0.048\ m

Current in the straight wire, i = 10.5 A

Let F is the magnitude of force is exerted on the wire. The magnetic force acting on an object of length l is given by :

F=ilB

F=10.5\ A\times 0.048\ m\times 0.52\ T

F = 0.262 N

So, the magnitude of force is exerted on the wire is 0.262 N.

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What would you say to a friend who made this statement, “The visible-light spectrum of the Sun shows weak hydrogen lines and str
Oliga [24]

Explanation:

spectral lines or signatures of elements depend on temperature, the temperature of the sun is about 5800 K.

at this temperature most calcium atoms are excited to higher energy states than hydrogen atoms and this means that calcium atoms are gonna have more signatures than the atoms of hydrogen.

the statement that the sun shows weak hyrogen lines and strong calcium line is wrong because at the sun's temperature most of the hydrogen atoms are in lower energy states while calcium atoms are in higher energy states hence calcium has more or ''strong'' lines than hydrogen.

8 0
3 years ago
3.00Kg toy falls from a height of 1.00m. What is the kinetic energy just before the ground?
ivanzaharov [21]

Answer:K E = 29.4 J

Explanation:

7 0
3 years ago
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
A laser pulse of duration 25 ms has a total energy of 1.4 J. The wavelength of this radiation is
SpyIntel [72]

Answer:

n = 4 x 10¹⁸ photons

Explanation:

First, we will calculate the energy of one photon in the radiation:

E = \frac{hc}{\lambda}\\\\

where,

E = Energy of one photon = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 567 nm = 5.67 x 10⁻⁷ m

Therefore,

E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{5.67\ x\ 10^{-7}\ m}

E = 3.505 x 10⁻¹⁹ J

Now, the number of photons to make up the total energy can be calculated as follows:

Total\ Energy = nE\\1.4\ J = n(3.505\ x\ 10^{-19}\ J)\\n = \frac{1.4\ J}{3.505\ x\ 10^{-19}\ J}\\

<u>n = 4 x 10¹⁸ photons</u>

8 0
3 years ago
Learning Goal: To understand the distinction between mass and weight and to be able to calculate the weight of an object from it
Nimfa-mama [501]

Answer:

The true statements are: A, D

Explanation:

This interesting problem of the conceptual relationship between mass and weight, the equation for weight is

        W = m g

From Newton's second law

       W = ma

Where g the acceleration of gravity, this acceleration can vary at several points, for example, in a vertical circumference the acceleration of gravity is always down and the centripetal acceleration continuously changes direction therefore the body weight constantly changes from zero to the maximum value.

The mass instead is always the same and is the resistance (inertia) to the movement of the bodies

Of the aforementioned the peo has the unit mass multiplied by the acceleration

           

        Weight [N] = mass [kg] acceleration [m / s2]

        Weight [lb] = mass [slug] acceleration [ft / s2]

Examine the statements

A) 12.0 lb. True pounds are the mass for acceleration. English measurement system

B) 0.34g False. Grams are units of mass,

C) 120 kg. False. The kilograms is a multiple of the grams, which are units of mass

D) 1600 kN True the newton is the unit of weight, the Newton kilo is a multiple

E) 0.34 m False meters are units of length

F) 411 cm False centimeters is a submultiple of the meter that is a unit of length

The true statements are: A, D

4 0
3 years ago
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