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svlad2 [7]
3 years ago
13

A mouse spots an owl flying 10.1 feet over a nearby bush. The mouse is 30 feet away from the bush, and it sees the owl at a cert

ain angle of elevation. Which trigonometric equation can be used to solve for x, the angle of elevation from the mouse to the owl?

Physics
1 answer:
Anna11 [10]3 years ago
6 0

Answer:

Explanation:

Given

fly is at height of h=10.1\ ft

mouse is at a distance of 30 ft from bush

if \theta  is the angle of elevation then

using trigonometric relation

\tan \theta =\frac{h}{x}

\tan \theta =\frac{10.1}{30}

\theta =tan^{-1}(\frac{10.1}{30})

\theta =18.57^{\circ}

           

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A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
An aircraft travels 1000. km west and then travels 500. km west. what is its total displacement?
UkoKoshka [18]
1000-500=500 km. to the west.
5 0
3 years ago
A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball
RideAnS [48]

Answer:

C=2.32\times 10^{-4}\ Ns^2/m^2

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, F_r=Cv^2

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

Cv^2=mg

C=\dfrac{mg}{v^2}

C=\dfrac{0.046\times 9.8}{(44)^2}

C=2.32\times 10^{-4}\ Ns^2/m^2

Hence, this is the required solution.

4 0
3 years ago
A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a
Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

F_N = F_C - F_G = m\frac{v^{2} }{r} - mg

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

E_{tot}=mgh + \frac{1}{2} mv^{2}

At the top of the hill:

E_{tot}= mgh_{hill}

At the top of the loop:

E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}

Setting both energies equal and canceling the mass m gives:

gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}

Solving for v:

v^{2} = 2g(h_{hill}-h_{loo}_p)

Using v in the first equation:

F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg

F_N = 955.5N

7 0
3 years ago
What is true about resistance? Check all that apply. A. It is the excess accumulation of electric charge B. It is measured in oh
Alekssandra [29.7K]

Answer:

B, C and E

Explanation:

The unit of resistance in the international system is the Ohm, the equation that describes the resistance is:

R=p\frac{l}{S} \\

Where (l) is for lenght of the wire, (S) is the area and (p) its the constant associated to the conductor.

It's related by the Ohm's Law:

R=\frac{V}{I}

3 0
3 years ago
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