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3 years ago

15

Our sun has a low absolute magnitude compared to many other stars but appears very bright to us because of its

1 answer:

nadya68 [22]3 years ago

6 0

relative closeness to earth

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Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

2 years ago

Read 2 more answers

If an astronaut throws an object in space, the object's speed will _____________. A) decrease. B) increase. C) remain constant.

GenaCL600 [577]

The object's speed will remain constant after the it leaves his hand.

So will HIS speed in the opposite direction.

7 0

2 years ago

Read 2 more answers

A frictionless pendulum is made with a bob of mass 19.7 kg. The bob is held at height = 0.934 meter above the bottom of its traj

Alika [10]

Answer:

265 J

Explanation:

$Energy=PE+KE=mgh+ 0.5mv^{2}$ where KE is kinetic energy, PE is potential energy, m is the mass of an object, v is the speed, h is the height and g is acceleration due to gravity.

Substituting 19.7 Kg for mass, 0.934 for h, 2.93 for v and 9.81 for g then

$Energy=19.7(9.81*0.934+0.5*2.93^{2})=265.063303\approx 265 J$

4 0

2 years ago

The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a

brilliants [131]

Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

vector (AB) = 2 i + j - 5 k

mag (AB) = sqrt (2^2 + 1^2 + 5^2)

mag (AB) = sqrt(30)

unit (AB) = ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

vector (T) = unit(AB)* 2.3 KN

= 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

vector (AC) = 2 i - 2 j - 5 k

mag (AC) = sqrt (2^2 + 2^2 + 5^2)

mag (AC) = sqrt(33)

unit (AC) = ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k) . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

T_ac = 0.2923947572 - 0.146973786 - 1.827467232

T_ac = 1.972888 KN

4 0

3 years ago

The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th

bagirrra123 [75]

The velocity of the wave on the string is given by

$v=\sqrt{\frac{T}{\frac{m}{L}}} \\ v=\sqrt{\frac{TL}{m}}$

Solving the above equation,

$v^2=\frac{TL}{m} \\ L=\frac{v^2m}{T}$

The frequency of the wave $f=300$ and wave length is $0.76$

The velocity is $v=(300)(0.76)=228$

Substituting numerical values,

$L=\frac{228^2(0.0059)}{200}\\ T=1.534$

The length of the string is $1.534 m$

4 0

3 years ago