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Tema [17]
2 years ago
14

A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are

oriented parallel to a uniform magnetic field of strength 0.037 T, the other two sides being perpendicular to the magnetic field.
A)What is the magnitude of the magnetic moment of the loop?
B)What torque does the magnetic field exert on the loop?
Physics
1 answer:
Dahasolnce [82]2 years ago
4 0

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

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Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

R = \frac{u^{2}Sin2\theta }{g}

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

R = \frac{u^{2}Sin60 }{9.8}

R = 0.088u^2

If the angle of projection is 45 degree.

R = \frac{u^{2}Sin90 }{g}

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5 0
2 years ago
Tom and his little sister are enjoying an afternoon at the ice rink. they playfully place their hands together and push against
Westkost [7]
Newton's third law says:
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So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>F_t = F_s
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".

From Newton's second law, we also know
</span>F=ma
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>m_t a_t = m_s a_s
<span>And find Tom's acceleration:
</span>a_t =  \frac{m_s}{m_t} a_s =  \frac{15 kg}{61 kg} (2.1 m/s^2)  =0.52 m/s^2<span>
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5 0
2 years ago
What best describes myotibrils
krok68 [10]
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8 0
1 year ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
horrorfan [7]

Answer:

The value is t_1 =  9 \  s

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

v  =  u   +  at

Here u = 0 \ m/s

a = 13.5 m/s^2

So

v  =  0   +  13.5 *  3.10

=> v  =  41.85 \ m/s

The is distance it covers at this time is

s =  u *  t  +  \frac{1}{2} a *  t^2

=> s =  +  \frac{1}{2} * 13.5 *  3.10^2

=> s =64.87

Now when sled stops its the final velocity is v_f =  0 m/s while the initial velocity will be the velocity after its acceleration i.e v  =  41.85 \ m/s

So

v_f  =  v  +  a_1t_1

Here  a_1 =  - 4.65, the negative sign shows that it is deceleration

So

           0  =  41.85  - 4.65 *  t_1

=> t_1 =  9 \  s

3 0
1 year ago
A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h
Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

mv_o=m(v_1 + v_2) where m is the mass, v_o is the initial speed before impact, v_1 and v_2 are velocity of the impacting object after collision and velocity after impact of the originally constant object

5m=m(v_1 +v_2)

Therefore v_1+v_2=5

After collision, kinetic energy doubles hence

2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})

2v_o^{2}=v_1^{2} + v_2^{2}

Substituting 5 m/s for v_o then

2*(5^{2})= v_1^{2} + v_2^{2}

50= v_1^{2} + v_2^{2}

Also, it’s known that v_1+v_2=5 hence v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

2v_2^{2}-10v_2-25=0

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then v_2=6.83 m/s

Substituting, v_1=-1.83 m/s

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0
1 year ago
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