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finlep [7]
3 years ago
9

Right hand rule exercise

Physics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer:

A yxiigxih5dd8yixc99uf8g

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A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal.
Korvikt [17]

Answer:

B) 27.3 m

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x =  vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 30 m/s , at an angle  α=40.0° above the horizontal

v₀x = vx = 30*cos40° = 22.98 m/s

v₀y = 30*sin40° = 19.28 m/s

y₀ = 2m

y =  18.0 m

g = 9.8 m/s²

Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

(vfy)² = 58.26

v_{f} = \sqrt{58.26}

vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

Horizontal distance from where the rock was thrown to the window

We replace t = 1.19 s , in the equation (1)

x =  vx*t  

x = (22.98)* ( 1.19 )

x = 27.3 m

3 0
3 years ago
The children at the local park think the slide is too slow. There is too much fiction. What could you do to decrease the amount
igor_vitrenko [27]
In order to decrease the friction on the slide,
we could try some of these:

-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide.  The oil oozes
down the slide and keeps the whole slide greased.

-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.

-- Spray the whole slide with soapy sudsy water, every 30 minutes.

-- Drill a million holes in the slide,and pump high-pressure air
through the holes.  Make the slide like an air hockey table.

-- Keep the slide very cold, and keep spraying it with a fine mist
of water.  The water freezes, and a thin coating of ice stays on
the slide.

-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide.  If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.

-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide.  Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.
7 0
3 years ago
how does the frequency of a radio wave compare to the frequency of the vibrating electrons that produce it?
HACTEHA [7]

Answer:

mass

Explanation:

3 0
3 years ago
Which of the following are functions of both cells and living organisms?
Nana76 [90]
I believe it is, All of the above.
6 0
3 years ago
A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis
romanna [79]

Answer:

2 rad/s

Explanation:

For a rotating object, the linear velocity is given by

v=\omega r

where \omega is the angular velocity and r is the radius.

\omega=\dfrac{v}{r}

The edge has a linear velocity of 10 m/s and the radius at the edge is 5 m.

\omega=\dfrac{10 \text{ m/s}}{5 \text{ m}} = 2 \text{ rad/s}

8 0
4 years ago
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