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3 years ago

Where is the center of gravity if the 9.00 kg mass of the barbell itself is taken into account?

Physics

1 answer:

Pavel [41]3 years ago

3 0

The problem is missing some parts but nevertheless here is the answer:

Given:

1.7 m long barbell with 23 kg weight on the left and 34 in the right end.

Solution:

Calculating from the left side:

Center of gravity = ΣM/ΣW = [23*0 + 9*(1.7/2) + 34*1.7]/[23+9+34]

Center of gravity = .992 m from the left end

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A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie

Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $$\theta$

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $$\theta$

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $$\theta$

U = Iπ $r^{2}$ B cos $$\theta$

For $$\theta$ = 0° →

U(Max) = MB cos(0) = MB = Iπ $r^{2}$ B = 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

375 π x $10^{-7}$ .

For $$\theta$ = 90° →

U = MB cos (90) = 0

For $$\theta$ = 180° →

U(Min) = MB cos(0) = - MB = - Iπ $r^{2}$ B = - 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

- 375 π x $10^{-7}$ .

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

#SPJ4

3 0

1 year ago

Giving 20 points and brainiest <br> But please help me

monitta

Answer:

yeah, Do you want me to check your answers? Yes is correct

8 0

3 years ago

What is a

LiRa [457]

Answer:

I think there six points

4 0

2 years ago

Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t

Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

$V_{b} - V_{a}$ = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

V_{b} - V_{a} = - ∫ E ds

We substitute

V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

V_{b} - V_{a} = - V_{b} - V_{a}

V_{a} - V_{b} = 89.3

3 0

3 years ago

Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l

Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1 (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2 (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2) (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5 (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2 (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

$R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067} \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22$

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

= 0.5 - 0.9931 × 0.340

= 0.5 - 0.338

= 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0

3 years ago