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Monica [59]
1 year ago
5

A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.

Physics
1 answer:
Kazeer [188]1 year ago
4 0

Answer:

(a) Magnetic moment will be 17.212\times 10^{-4}A-m^2

(b) Torque will be 6.024\times 10^{-4}N-m

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2

Current is given as i=15mA=15\times 10^{-3}A

Number of turns N = 25

(A) We know that magnetic moment is given by magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by \tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m

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A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,
Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

d_x = 10 + 2= 12 blocks

Total displacement towards South

d_y = 5 block

now to find the net displacement we can use vector addition

d = \sqrt{d_x^2 + d_y^2}

d = \sqrt{12^2 + 5^2}

d = 13 blocks

<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0
1 year ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
2 years ago
A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the
Blizzard [7]
<h3><u>Answer;</u></h3>

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant  

F = k*y  

Therefore;  k = F/y = m*g/y

                               = 0.40kg*9.8m/s^2/(0.76 - 0.41)

                               = 11.2 N/m  

Energy is conserved  

Let A be the maximum displacement  

Therefore;  1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2  

Thus;  A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

               = sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

                = 0.846 m  

Thus; the length will be 0.41 + 0.846  = 1.256 m

6 0
2 years ago
A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?
Talja [164]

Answer:

1.05 N

Explanation:

K = 0.7 N/m

e = 1.5 m

F = ?

from Hooke's law of elasticity

F = Ke

= 0.7×1.5

= 1.05 N

5 0
2 years ago
The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2
masha68 [24]

Answer:

the normal stress induced by  the concrete post \sigma_c = 67.26 psi

the normal stress induced by the steel \sigma_s = - 1795.84 psi

Explanation:

Given that:

Modulus for elasticity for concrete post E_c = 3.6 *10^6 psi

Thermal coefficient for concrete post  \alpha _c = 5.5 *10^{-6}/^0F

Modulus for elasticity of steel bar E_s = 29*10^6psi

Thermal coefficient of steel bar \alpha _2 = 6.5*10^{-6}/^0F

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod A_s = 6(\frac{ \pi}{4} )(d_s)^2

= 6(\frac{ \pi}{4} )(\frac{7}{8} )^2

= 3.61 in²

Area of concrete parts A_c = (10)(10) - A_s

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= [\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)

= [\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)

= [(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}

= 1.234*10^{-8}P = 8.0*10^{-5}

P = \frac {8.0*10^{-5}}{1.234*10^{-8}}

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

\sigma_c =\frac{P}{A_c}

\sigma_c = \frac{6482.98}{96.39}

\sigma_c = 67.26 psi

Thus, the normal stress induced by  the concrete post \sigma_c = 67.26 psi

Also; the normal stress in the steel bars  induced as a result of temperature rise is as follows:

\sigma_s = \frac{-P}{A_s}

\sigma_s =\frac{-6482.98}{3.61}

\sigma_s = - 1795.84 psi

Thus, the normal stress induced by the steel \sigma_s = - 1795.84 psi

6 0
2 years ago
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