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3 years ago

15

If the vertical posts supporting these wires divide the lines into straight 19.0 m segments, what magnetic force does each segme

nt exert on the other?

1 answer:

erica [24]3 years ago

5 0

Answer:

Complete question is:

Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each other 38.0 cm apart.

Part A.) If the vertical posts supporting these wires divide the lines into straight 20.0m segments, what magnetic force does each segment exert on the other? (answer in F= ? N)

Part B.) Is the force found in part (A) attractive or repulsive?

Answer: 0.024 N attraction force is being applied.

Explanation:

See attached picture.

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Differentiate between center of mass and center of gravity

ELEN [110]

Answer:

Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.

Explanation:

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3 0

3 years ago

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Help immediately! Answer I need help.

garri49 [273]

Answer:

0.80 m

Explanation:

elastic potential energy formula

elastic potential energy = 0.5 × spring constant × (extension) 2

4 0

2 years ago

A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pres

Vikentia [17]

Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

V = L^3 = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

P*V = n*R*T

n = P*V / R*T

n = 101.3*10^3 * 0.001 / 8.314*300

n = 0.04061 moles

- Compute the mass of the gas:

m = n*M

m = 0.04061*28.9

m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

F_g = m*g

F_g = 1.174*9.81*10^-3

F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

F_c = P*A

F_c = (101.3*10^3)*(0.1)^2

F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

4 0

3 years ago