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2 years ago

I will mark you brainlist!

Physics

1 answer:

kirill115 [55]2 years ago

7 0

Tornado- Trees knocked down, debris everywhere, ground and dirt scattered.

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A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie

ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, $A=7.4\ cm^2=7.4\times 10^{-4}\ m^2$

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, $B_1=0.5\ T$

Final magnetic field, $B_2=3\ T$

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{B_2-B_1}{t}$

$\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}$

$\epsilon=1.65\times 10^{-3}\ V$

Induced current in the loop of wire is given by :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{1.65\times 10^{-3}}{2.4}$

$I=6.87\times 10^{-4}\ A$

So, the induced current in the loop of wire over this time is $6.87\times 10^{-4}\ A$ . Hence, this is the required solution.

7 0

2 years ago

A thunderstorm loses energy in the _______ stage.

Irina-Kira [14]

dissipation is the answer ;(

3 0

2 years ago

Read 2 more answers

The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

2 years ago

please help me guys please please please please please please please please please please please please please

seropon [69]

Answer:

1.F = 256 N

2.a = 8 m/s/s

By looking at the given information, you know force, and an acceleration. Therefore you have enough information to use the first formula.

F = ma

256 N = m * 8 m/s/s

m = 256 N/8 m/s/s

m = 32 Kg

6 0

2 years ago

Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite

nalin [4]

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

$\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s$

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

$J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s$

Hence, this is the required solution.

7 0

2 years ago