Ask question

Ask question

All categories

3 years ago

10

The bearing of a lighthouse from a ship was found to be N 37 degrees E. After the ship sailed 2.5 miles due south, the new beari

ng was N 25 degrees East. Find the distance between the ship and the lighthouse at each location.

1 answer:

ella [17]3 years ago

5 0

Answer:

The distance between the lighthouse and the ship

from the start position A = 5.08 miles

from the Final point B = 7.23 miles

Explanation:

Note: Refer the figure

Let the position of the lighthouse be 'L'

Given:

When the ship is at the position A, ∠DAL=37°

Now, when the ship sails through a distance of 2.5 i.e at position B

mathematically,

AB=2.5 miles

∠ABL=25°

Now,

∠DAL + ∠LAB = 180°

or

37° + ∠LAB = 180°

or

∠LAB = 180° - 37° = 143°

Also, In ΔLAB

∠LAB + ∠ABL + ∠ALB = 180°

or

143° + 25° + ∠ALB = 180°

or

∠ALB = 180° - 143° - 25° = 12°

Now using the concept of the sin law

In ΔLAB

$\frac{AL}{sin25^o}=\frac{2.5}{sin12^o}$

or

AL = 5.08 miles

and,

$\frac{BL}{sin143^o}=\frac{2.5}{sin12^o}$

or

BL = 7.23 miles

hence,

The distance between the lighthouse and the ship

from the start position A = 5.08 miles

from the Final point B = 7.23 miles

You might be interested in

how do you solve this problem. A bat flying at 3.7m/s spots an insect 23.8 meters away. How much must the bat increase its speed

RideAnS [48]

Here's how I would do it:

How far does he have to go to catch the bug ? <u>23.8 meters</u>

How soon does he want to get there ? <u>1.8 seconds</u>

What speed does he need ? (23.8 m) / (1.8 sec) = <u>13.222 m/sec</u>

What speed is he flying now? <u>3.7m/s</u>

How much does he need to increase it ? (13.222 - 3.7) = <em>9.5 m/s</em> faster

5 0

4 years ago

A jet engine pushes air at high velocity toward the rear of the plane. Thrust is created moving the plane forward. Which stateme

Lapatulllka [165]

The air pushes the engine with an equal force in the opposite direction.
Whenever force is used there is an opposite yet equal affect.

5 0

3 years ago

If two light waves are coherent a) their amplitudes are the same their phase difference is constant their frequencies are the sa

Aleonysh [2.5K]

Answer:

a) their amplitudes are the same their phase difference is constant their frequencies are the same

Explanation:

Coherent waves are the waves that have constant phase difference, equal frequency, amplitude and waveform.

Frequency denotes the number of cycles a wave completes in one second.

Amplitude is the maximum height that the wave reaches.

Waveform is the two dimensional representation of a wave in graphical form.

3 0

3 years ago

A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He

Taya2010 [7]

a) 73.5 N

b) 551.3 J

c) -551.3 J

d) 0 J

e) 0 J

f) 0 J

g) 0 J

Explanation:

a)

There are two forces acting on the crate:

- The push of the worker, F, in the forward direction

- The frictional force, $F_f=\mu mg$ , in the backward direction, where

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:

$F-F_f=ma$

However, the crate here is moving with constant velocity, so its acceleration is zero:

$a=0$

So the previous equation becomes:

$F-F_f=0$

And we can find the magnitude of the applied force:

$F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N$

b)

The work done by the applied force on the crate is

$W_F=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here we have:

F = 73.5 N

d = 7.5 m

$\theta=0^{\circ}$ (the force is applied in the same direction as the displacement)

Therefore,

$W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J$

c)

The work done by friction on the crate is:

$W_{F_f}=F_f d cos \theta$

where in this case:

$F_f=73.5 N$ is the magnitude of the force of friction

d = 7.5 m is the displacement of the crate

$\theta=180^{\circ}$ , because the displacement is forward and the force of friction is backward, so they are in opposite direction

Therefore, the work done by the force of friction is:

$W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J$

d)

To find the normal force, we analyze the situation of the force along the vertical direction.

We have two forces on the vertical direction:

- The normal force, N, upward

- The force of gravity, $mg$ , downward, where

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:

$N-mg=0\\N=mg=(50)(9.8)=490 N$

The work done by the normal force is:

$W_N=Nd cos \theta$

In this case, $\theta=90^{\circ}$ , since the normal force is perpendicular to the displacement of the crate; therefore, the work done is

$W_N=(490)(7.5)(cos 90^{\circ})=0$

e)

The work done by the gravitational force is:

$W_g=F_g d cos \theta$

where:

$F_g=mg=(50)(9.8)=490 N$ is the gravitational force

d = 7.5 m is the displacement of the crate

$\theta=90^{\circ}$ is the angle between the direction of the gravitational force (downward) and the displacement (forward)

Therefore, the work done by gravity is

$W_g=(490)(7.5)(cos 90^{\circ})=0 J$

f)

The total work done on the crate can be calculated by adding the work done by each force:

$W=W_F+W_{F_f}+W_N+W_g$

where we have:

$W_F=+551.3 J$ is the work done by the applied force

$W_{F_f}=-551.3 J$ is the work done by the frictional force

$W_N=0$ is the work done by the normal force

$W_g=0$ is the work done by the force of gravity

Substituting,

$W=+551.3+(-551.3)+0+0=0 J$

So, the total work is 0 J.

g)

According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:

$W=\Delta E_K$

where

W is the work done on the crate

$\Delta E_K$ is the change in kinetic energy of the crate

In this problem, we have:

$W=0$ (total work done on the crate is zero)

Therefore, the change in kinetic energy of the crate is:

$\Delta E_K = W = 0$

5 0

4 years ago

How does the size of arctic foxes ears help to keep them warm in a cold environment

DochEvi [55]

artic foxes have thick white fur to keep them warm and also for camouflage in the snow they have small ears to keep the snow out of their ears so that they can hear better they also have thick fur lining and a thick pad on their feet this stops the paws from getting cold and they have short legs so then the body weight can be spread making it easier for the animal to not sink in the snow in the summer the artic fox turns brown to help make the fox invisible when people are hunting in the long grass

3 0

3 years ago