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3 years ago

A copper telephone wire has essentially

Physics

1 answer:

Lunna [17]3 years ago

5 0

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

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Two boxes on a horizontal plane with coefficient of friction µ are connected by a massless string. The left-hand box has mass m

pochemuha

Answer:

T = (3μmg - Fcosθ)/2

Explanation:

Since the boxes move to the right, the net force on box of mass 2m is

Fcosθ - 2μmg + T = ma (1)where Fcosθ = horizontal component of applied force, 2μmg = frictional force and T = tension in string and a = acceleration of box

The net force on the box with mass m is

μmg - T = ma (2) where μmg = frictional force, T = tension and a = acceleration of boxes.

Equating (1) and (2) we have

Fcosθ - 2μmg + T = μmg - T

collecting like terms

Fcosθ - 2μmg - μmg = -T - T

Fcosθ - 3μmg = -2T

(3μmg - Fcosθ)/2 = T

T = (3μmg - Fcosθ)/2

5 0

3 years ago

While painting the top of an antenna 275 m in height, a worker accidentally lets a 1.00 L water bottle fall from his lunchbox. T

Tom [10]

Answer:

The force of impact of the water bottle is F = 13,475 N

Explanation:

Given data,

The height of the antenna, h = 275 m

The mass of the 1 L water bottle, m = 1 kg

Let the bottle moves distance immediately after the impact is, d = 0.2 m

The force exerted by the bottle on the bushes at the ground is given by the formula,

F = mgh / d

Substituting the values

F = 1 x 9.8 x 275 / 0.2

= 13,475 N

The value of the force of impact can be reduced by increasing the value of d, it is like the lowering the hand along with the motion of the ball to catch it thereby reduce the force of impact.

The force of impact of the water bottle is F = 13,475 N

8 0

3 years ago

What is a circuit,s resistance if 12v produces 2A of current ?

Arlecino [84]

We have to find the resistance.
V = 12
A = 2
R = ?
And then we divide it to find r.
12/2 = 6 ohm's.

7 0

3 years ago

As you stand near a railroad track, a train passes by at a speed of 31.7 m/s while sounding its horn at a frequency of 218 Hz. W

Darya [45]

Explanation:

Given that,

Frequency of train horn, f = 218 Hz

Speed of train, $v_t = 31.7 m/s$

The speed of sound, V = 344 m/s (say)

The speed of the observed person, $V_o=0\ m/s$

(a) When the train approaches you, the Doppler's effect gives the frequency as follows :

$f'=f(\dfrac{V}{V-v_t})\\\\f'=218\times (\dfrac{344}{344-31.7})\\\\f'=240.12\ Hz$

(b) When the train moves away from you, the Doppler's effect gives the frequency as follows :

$f'=f(\dfrac{V}{V+v_t})\\\\f'=218\times (\dfrac{344}{344+31.7})\\\\f'=199.6\ Hz$

Hence, this is the required solution.

6 0

4 years ago

The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;