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3 years ago

7

A sailboat runs before the wind with a constant speed of 2.8 m/s in a direction 52° north of west. How far (a) west and (b) nort

h has the sailboat traveled in 35 min?

1 answer:

vodka [1.7K]3 years ago

6 0

<h2>Displacement along west = 3612 m</h2><h2>Displacement along north = 4633.50 m</h2>

Explanation:

Let east be positive x axis and north be positive y axis

Velocity of boat = 2.8 m/s in a direction 52° north of west.

Velocity, v = -2.8 cos 52 i + 2.8 sin 52 j = -1.72 i + 2.21 j m/s

Time taken = 35 min = 35 x 60 = 2100 s

Displacement = Velocity x Time

Displacement = (-1.72 i + 2.21 j) x 2100

Displacement = -3612 i + 4633.50 j m

Displacement along west = 3612 m

Displacement along north = 4633.50 m

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Emmett is lifting a box vertically. Which forces are necessary for calculating the total force?

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When Emmett is lifting a box vertically, the forces that must be added to calculate the total force are: the gravitational force, tension force(the force exerted by Emmett to the box and the force exerted by the box to Emmett), and air resistance force.

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The number of kilowatts used by an individual to operate his appliances is determined as 12.1 kWh.

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2 years ago

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3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

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$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

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$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

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And the velocity in the constricted section is

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3 years ago

The resistance (R) of a copper wire varies directly as its length (L). Write this relation as a formula using k as the constant

Drupady [299]

Answer:

Explanation:

According to ohm's law current flowing in a conductor is directly proportional to the voltage applied across two end of conductor.

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whee L and d are length and Diameter

thus $R=k \frac{L}{d^2}$

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3 years ago