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Charra [1.4K]
3 years ago
9

a wire carrying 1.5 A passes through a 48 mT magnetic field the wire is peremndicular to the field and makes a quarter circle tu

rn radius 21 cm in the dield regeion. Find the magnitude and direction of the magnetic force
Physics
1 answer:
professor190 [17]3 years ago
8 0

Answer:

Explanation:

Wire is in the form of an arc of the size of a quarter of a circle. The distance between its two end

=√2 x 21 cm

= 29.7 cm

= 29.7 x 10⁻² m

Force on the arc in the magnetic field will be same as the straight line meeting its two ends.

Force = BIL

= 48 x10⁻³ x  1.5 x 29.7 x 10⁻²

= 2138.4 x 10⁻⁵ N.

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Which statement(s) accurately describe the conditions immediately before and after the firecracker explodes:
lidiya [134]

Answer:

Option C, The total momentum of the fragments is equal to the original momentum of the firecracker.

Explanation:

Kinetic energy of cracker cannot remain constant before and after explosion. It is so because in the process of burning and bursting some amount of kinetic energy is lost in the form of light and heat energy. While the total mass before and after the explosion remains constant due to which the momentum is conserved before and after the explosion

Hence, option C is correct

8 0
3 years ago
A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.1
Shkiper50 [21]

Answer:

0.10013 atm

Explanation:

Applying Boyle's Law,

P'V' = PV................... Equation 1

Where P' = Initial pressure of air, V' = Initial volume of air, P = Final pressure of air, V = Final volume of air.

make P the subject of the equation

P = P'V'/V..................... Equation 2

Given: P' = 0.355 atm, V'  0.110 m³, V = 0.390 m³

Substitute into equation 2

P = 0.355(0.11)/0.39

P = 0.10013 atm.

5 0
3 years ago
Which BEST describes the speed and velocity of the graph?
Neko [114]
I think the answer is A (sorry if it isn't).
6 0
3 years ago
Read 2 more answers
Prove that..<br>please help<br>​
GaryK [48]

\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

Also,

\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

5 0
2 years ago
Read 2 more answers
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