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marishachu [46]

2 years ago

10

A mail carrier leaves the post ofﬁce and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

km. After dropping off a package, he drives 9.50 km 35.0° north of east to Starbucks. What is the direction relative to the positive x axis?

1 answer:

den301095 [7]2 years ago

7 0

Answer:

$\theta=7^o$

Explanation:

<u>Displacement</u>

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

$\vec r_1=\ km$

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

$\vec r_2=\ km=\ km$

Finally, he drives 9.5 km 35° north of east.

$\vec r_3=\ km=\ km$

The total displacement is

$\vec r_t=\ km+\ km+\ km$

$\vec r_t=\ km$

The direction can be calculated with

$\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232$

$\boxed{\theta=7^o}$

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13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?

sergeinik [125]

The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

Explanation:

The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

$Q = mc del T$

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

$Q = 20*C*80$

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

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2 years ago

A wheel completes 5.6 revolutions in 8 seconds.

nevsk [136]

Answer:

$86.15\pi rad/min$

Explanation: Angular velocity is the number of revolutions made per unit time.

We convert the number of revolutions to radians and the time given in seconds to minutes,

Given;

$1rev=2\pi rad\\therefore\\5.6rev=5.6*2\pi rad\\= 11.2\pi rad$

Also,

60s = 1 min

hence

$8s=\frac{8}{60}min\\=0.13min$

We now divide the number of revolution in radians by the time in minutes.

$\omega =\frac{11.2\pi}{0.13min}\\\omega=86.15\pi rad/min$

5 0

2 years ago

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If you want to maximize the magnetic force on a current in a conductor, how should you orient the current relative to the magnet

tia_tia [17]

Answer:

Should place the current perpendicular to the magnetic field

Explanation:

The magnetic force exerted on a current-carrying wire is given by

$F=ILBsin \theta$

where

I is the current in the wire

L is the length of the wire

B is the magnetic field

$\theta$ is the angle between the direction of the wire and the magnetic field

As we see from the formula, the magnetic force is maximum when

$sin \theta=1$

which means

$\theta=90^{\circ}$

So, when the current in the wire and the magnetic field are perpendicular to each other.

7 0

2 years ago

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

Tema [17]

Answer:

The value of developed electric force is $3.516\times 10^{- 7} N$

Solution:

As per the question:

Mass of the droplet = 1.8 mg = $1.8\times 10^{- 6} kg$

Charge on droplet, Q = $25 pC = 25\times 10^{- 12} C$

Distance between the 2 droplets, D = 0.40 cm = 0.004 m

Now, the Electrostatic force given by Coulomb:

$F_{E} = \frac{1}{4\pi epsilon_{o}}.\frac{Q^{2}}{D^{2}}$

$\frac{1}{4\pi epsilon_{o}} = 9\times 10^{9} m/F$

$F_{E} = (9\times 10^{9}).\frac{(25\times 10^{- 12})^{2}}{0.004^{2}}$

$F_{E} = 3.516\times 10^{- 7} N$

The magnitude of force is too low to be noticed.

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2 years ago

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A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410,

Alik [6]

The horizontal pushing force required to just start the crate moving is 447 N.

The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

<u>Explanation</u>:

By the definition of the coefficient of static friction we have:

μ $_{s}$ = $\frac{F_{appl} }{W}= \frac{F_{s} }{N}$ ,

where, $F_{appl}$ is the horizontal pushing force,

W = mg is the weight of the crate directed downward,

$F_{s}$ is the static friction force-directed opposite to the horizontal pushing force and equal to it,

N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to just start the crate moving:

$F_{appl} = F_{s} = u_{s}N = u_{s}mg$

= 0.760 $\times$ 60 kg $\times$ 9.8 m / s^2

= 447 N.

By the definition of the coefficient of kinetic friction we have:

u $_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N}$ ,

where, $F_{appl}$ is the horizontal pushing force,

W = mg is the weight of the crate directed downward,

$F_{k}$ is the kinetic friction force-directed opposite to the horizontal pushing force and equal to it,

N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to slide the crate across the dock at a constant speed:

$F_{appl} = F_{k} = u_{k}N = u_{k}mg$

= 0.410 $\times$ 60 $\times$ 9.8

= 241 N.

The horizontal pushing force required to just start the crate moving is 447 N.

The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

6 0

2 years ago