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lara31 [8.8K]
3 years ago
13

A mountaintop is a height y above the level ground. A woman measures the angle of elevation of the mountaintop to be θ when she

is a horizontal distance x from the mountaintop. After walking a distance d closer to the mountain, she measures the angle of elevation of the mountaintop to be φ. Neglecting the height of the woman’s eyes above the ground, draw a well-labeled diagram representing this situation and find an expression for the height of the mountain, y, in terms of d, φ, and θ. Note that your expression cannot contain x.
Physics
1 answer:
valentina_108 [34]3 years ago
7 0
Let t = Theta and p = Phi
Tan t = y/x    Then x =y/Tant.
Tant = y/(x-d)  x-d = y/Tanp
y/Tant  - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp  =  d*Tanr
y(1 - Tanr/Tanp = d*Tant

y = d*Tant/(1-Tant/Tanp)
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f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

h(t)=35t-0.83t^2

We know that the rate of change of displacement is equal to the velocity of an object.

v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t

Velocity of the arrow after 3 seconds will be :

v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0
3 years ago
according to isaac newtons theory of gravitation the force exerted between two objects is dependent on- A- an objects weight. B-
puteri [66]
A an objects weight hope this helps! :D 
3 0
3 years ago
What is the distance covered in 20 minutes by a train traveling 500m/m?
MrMuchimi

Answer:

answer is 10km

Explanation:

use "S =Ut "

S=distance U=velocity t =time

no need to convert time into seconds as the velocity has given in meters per minute

3 0
3 years ago
A traveler pulls on a suitcase strap at an angle 36° above the horizontal. If of work are done by the strap while moving the sui
ollegr [7]

Answer:

The tension in the strap is 74.82 N.

Explanation:

Given that,

Angle between the horizontal and the suitcase is 36 degrees.

The distance traveled by the suitcase is 15 meters.

Let the work done by the suitcase is 908 J. We know that the work done in the vector form is given by :

W=Fd\ \cos\theta\\\\908=F\times 15\times cos(36)\\\\F=74.82\ N

So, the tension in the strap is 74.82 N. Hence, this is the required solution.

6 0
3 years ago
Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t
GenaCL600 [577]
B) 14.0 N    

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is: 

 E = 0.5 M V^2 

 where 

 E = Energy 

 M = Mass 

 V = velocity   

 Substituting the known values, let's calculate the before and after energy. 

 Before: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (2.25 m/s)^2 

 E = 6.75 kg 5.0625 m^2/s^2 

 E = 34.17188 kg*m^2/s^2 = 34.17188 joules   

 After: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (1.2 m/s)^2 

 E = 6.75 kg 1.44 m^2/s^2 

 E = 9.72 kg*m^2/s^2 = 9.72 Joules   

 So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.   

 24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N   

 Rounding to 1 decimal place gives 14.0 N which matches option "B".
8 0
3 years ago
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