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AlexFokin [52]
3 years ago
8

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the vertical velocity compone

nt of the ball
Physics
1 answer:
sp2606 [1]3 years ago
3 0
Hello,

We have the vectorial formula

V = Vxi + Vyj

Where,

Vx = v.Cos(x)

And

vy = v.Sen(x)

We know too:

x = 35°

And

v = 31 m/s

Then we will have:

Vy = v.Sen(x)j

Vy = 31.Sen(35)j

Vy = (17,78 m/s) j
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Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

a= -9.81 m/s^2 t=???

Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

-15mx2/a= t^2

Square root (-30/-9.81m/s^2)

t=1.75 s

b) now v2!!

Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

v2 = -17.2 m/s or you can say 17.2 m/s down!!!

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2 years ago
A 0.50-kg croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it i
NeTakaya

The impulse given to the ball is equal to the change in its momentum:

J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s

This is also equal to the product of the average force and the time interval ∆t :

J = F(ave) ∆t

so that if F(ave) = 200 N, then

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6th grade science I mark as brainliest
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Answer:

8. organelle

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The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v
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Answer:

\theta = n\pi/2, {\rm where~n~is~an~integer.}

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k

Similarly,

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Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)

So,

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0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}

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Explaination: because its final.
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3 years ago
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