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slavikrds [6]
4 years ago
11

UPVOTES FOR EVERY ANSWER!!!!!!!!

Physics
2 answers:
r-ruslan [8.4K]4 years ago
6 0

Answer:

Option (c) is correct.

Explanation:

There are some properties of magnetic field lines which are given below:

(i) The field lines begin near the magnetic north pole and extend towards the south pole of the magnet outside the magnet.

(ii) Inside the magnet, the magnetic filed lines moves from south pole to north pole.

(iii) The tangent at any point on the field line gives the direction of magnetic field at that point.

mario62 [17]4 years ago
4 0
<span>C. Field lines begin near the magnet’s north pole and extend toward its south pole. 

Hope this helps!</span>
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A locomotive is pulling 15 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (w
VMariaS [17]

Answer:

298,220 N

Explanation:

Let the force on car three is T_23-T_34

Since net force= ma

from newton's second law we have

T_23-T_34 = ma

therefore,

T_23-T_34 = 37000×0.62

T_23= 22940+T_34

now, we need to calculate

T_34

Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2

F= ma

So, F= 12×37000×0.62= 22940×12= 275280 N

T_23 =22940+T_34= 22940+ 275280= 298,220 N

therefore, the tension in the coupling between the second and third cars

= 298,220 N

3 0
3 years ago
A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic
ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

I_i = Initial angular momentum = 275\ kgm^2

\omega_i = Initial angular velocity = 14 rev/min

I_f = Final angular momentum

\omega_f = Final angular velocity

Here the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s

The final angular velocity is 0.2932 rad/s

7 0
3 years ago
A planet has two small satellites in circular orbits around the planet. the first satellite has a period 18.0 hours and an orbit
puteri [66]
The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:
\frac{r_1^3}{T_1^2}= \frac{r_2^3}{T_2^2}
where
r_1 is the orbital radius of the first satellite
r_2 is the orbital radius of the second satellite
T_1 is the orbital period of the first satellite
T_2 is the orbital period of the second satellite

If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite:
T_2 =  \sqrt{T_1^2 ( \frac{r_2}{r_1} )^3} = \sqrt{(18.0 h)^2 ( \frac{3\cdot 10^7 m}{2 \cdot 10^7 m} )^3}  = 33.1 h
8 0
3 years ago
You are planning to make an open rectangular box from an 8-in.-by-15-in. piece of cardboard by cutting congruent squares from th
mariarad [96]
Should be 4x5x4!! If you take 8•15, you get 120!! There are six sides on a rectangular prism!! Thus leaving a value of 20 per side!! And I noted to properly equal out with the maximum volume it should mean 4•5=20!! Thus being 4•4•5
3 0
3 years ago
A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of
Alexandra [31]

Answer:

The maximum no. of electrons- 2.25\times 10^{22}

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

I = \frac{Q}{t}

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

Q = 1.0\times 3600 = 3600\ C

Maximum number of electrons, n is given by:

n = \frac{Q}{e}

where

e = charge on an electron = 1.6\times 10^{- 19}\ C

Thus

n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}

3 0
4 years ago
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