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2 years ago

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PLEASE HELPPP 100 POINTS HURRY !!!!Which diagram best illustrates the magnetic field of a bar magnet? A bar magnet with a north

and south pole. Field line vectors are leaving the north pole and curl around to the south pole. A bar magnet with a north and south pole. Field line vectors are leaving the south pole and curl around to the north pole. A bar magnet with a north and south pole. Field line vectors are leaving the north pole and curl around to the south pole. A bar magnet with a north and south pole. Field line vectors are leaving the south pole and curl around to the north pole.

2 answers:

Serggg [28]2 years ago

7 0

I think this is right I hope this is right for you

Iteru [2.4K]2 years ago

6 0

Answer:

A bar magnet with a north and south pole. Field line vectors are leaving the north pole and curl around to the south pole.

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

2 years ago

A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d

arsen [322]

Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7 , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional magnetic flux is induced out of the page.

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2 years ago

A motor boat is traveling east with a velocity of 7.3 meter/seconds. Its experiences a current of 0.34 meters/seconds from the e

andreev551 [17]

Since they're going in the same direction, just add the velocities together.

7.3 m/s + 0.34 m/s = 7.64m/s

3 0

2 years ago

Read 2 more answers

A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction

aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

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2 years ago

The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle

serg [7]

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we know that

$m_1 = 50 g = 0.050 kg$

$m_2 = 3 kg$

$v_{1i} = v_{2i} = 0$

$v_{1f} = 400 m/s$

now we will have

$0.050(0) + 3(0) = 0.050(400) + 3(v)$

$20 + 3v = 0$

$v = - \frac{20}{3} = - 6.67 m/s$

so gun will recoil with speed 6.67 m/s

6 0

2 years ago