Ask question

Ask question

All categories

3 years ago

9

While standing atop a building 49.6 m tall, you spot a friend standing on a street corner. Using a protractor and a dangling plu

mb bob, you find that the angle between the horizontal and the direction to the spot on the sidewalk where your friend is standing is 23°. Your eyes are located 1.74 m above the top of the building.
How far away from the foot of the building is your friend?

1 answer:

olga nikolaevna [1]3 years ago

6 0

Answer:

75degree don't forget wind and gravity force pulling down

You might be interested in

As the kinetic energy of an object increases, so does the _________ energy produced. A) heat B) light C) atomic D) potential

IrinaK [193]

Answer:

A) heat

Explanation:

As kinetic energy increases, so does the heat energy produced. The faster the molecules move, the more heat that is generated.

5 0

3 years ago

Read 2 more answers

What is the Sl unit of Newton ?............

kenny6666 [7]

Answer:

<em>N</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>of</em><em> </em><em>Newton</em>

5 0

3 years ago

Read 2 more answers

Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10

strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

$n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42$

So, the refractive index of the medium is 1.42.

5 0

3 years ago

What was the name of the voyages taken by Zheng he during the Ming dynasty on behalf of China

telo118 [61]

Answer:

Ming treasure voyages

Explanation:

6 0

3 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago