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3 years ago

Each graph has an x- and y-axis except

Physics

1 answer:

lianna [129]3 years ago

5 0

What are the choices?

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P.E

Vladimir79 [104]

Answer:

pagtalon?

Explanation:

because that had to be the answer

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3 years ago

How can you find the net force if two forces act in opposite directions?

Tresset [83]

Then the magnitude of the net force is the difference between the two forces,
and its direction is the same as the direction of the greater one.

4 0

3 years ago

Heredity appears to have a large effect on the intelligence of adults than environment. True False

Annette [7]

I JUST WANT THE ANSWER

5 0

3 years ago

Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it

MakcuM [25]

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r = r-h / r

2h = r-h

3h = r

h = r / 3

5 0

3 years ago

A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo

Alla [95]

Answer:

$v_{ox}= 19.6\ m/s$

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = $v_{ox}$

Initial vertical velocity= $v_{oy}$ (Since ball was thrown horizontally only)

Acceleration acting horizontally, $a_x$ = 0 m/s² [ Since no acceleration acts horizontally) ]

Vertical Acceleration, $a_y$ = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

$H= v_{oy}t+\frac{1}{2}a_yt^2$

$5= (0)t+\frac{1}{2}(9.8)t^2$

$t= \frac{10}{9.8}=1.02 s$

Then using Equations of motion for horizontal motion,

$R= v_{ox}t+\frac{1}{2}a_xt^2$

$20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2$

$v_{ox}= 19.6\ m/s$

4 0

3 years ago