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e-lub [12.9K]
3 years ago
10

A single-turn current loop, carrying a current of 3.50 a, is in the shape of a right triangle with sides 50.0, 120, and 130 cm.

the loop is in a uniform magnetic field of magnitude 72.5 mt whose direction is parallel to the current in the 130 cm side of the loop. what are the magnitude of the magnetic forces on each of the three sides?
Physics
1 answer:
dimulka [17.4K]3 years ago
3 0
Thank your very much
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Help me please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Vitek1552 [10]
C, due to stuff and whatnot
4 0
3 years ago
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A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will i
mafiozo [28]

Answer:

   t = 0.24 s

Explanation:

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation:  ΣF = ma

Rotation:      ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

                    ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation:     Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

         (1)                 mgSin27° - Ffr = ma  

For rotation:         (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

        (2)                Ffr = 2/3 ma

We substitute equation (2) in equation (1):

                            mgSin27° - 2/3 ma = ma

                            mgSin27° = ma + 2/3 ma

The mass gets cancelled:

                            gSin27° = 5/3 a

                            a = (3/5)(gSin27°)

                            a = (3/5)(9.8 m/s²(Sin27°))

                            a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

                           V = √2ad; where  d = 0.327m

                           V = √2(2.67 m/s²)(0.327m)

                            V = 1.32 m/s

Because V = d/t

                             t = d/V

                             t = 0.327m/1.32 m/s

                             t = 0.24 s

7 0
3 years ago
Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
A particle moving with initial velocity of 5m/s is subjected to a uniform acceleration of -2.5 m/s2.find the displacement for ne
LUCKY_DIMON [66]
1/2vt^2-1/2at^2
1/2*5*16-1/2*2,5*16=20
6 0
2 years ago
How can I shorten this and for it to sill make sense?
Julli [10]

Answer:

Radiation is the emission or transmission of energy in the form of waves.

Explanation:

.

8 0
3 years ago
Read 2 more answers
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