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maria [59]
3 years ago
14

Find the velocity in m/s of a baseball thrown 38m from third base to first base in 1.7s​

Mathematics
2 answers:
Mrrafil [7]3 years ago
5 0
V=s/t

Where V the velocity, S is the distance and T is time in SECONDS

V= 38/1.7

V≈ 22.4 ms -1
Oksi-84 [34.3K]3 years ago
3 0

Answer:

About 22m per second.

Step-by-step explanation:

All you have to do is 38/1.7

the answer is about 22

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An object is hanging by a string from your rearview mirror. While you are decelerating at a constant rate from 25 m/s to rest in
lidiya [134]
The rate of deceleration is equal to 25 m/s / 6 s = 25/6 m/s^2. 


(a) The string should be angled in such a way that the horizontal acceleration is equal to 25.6 and the vertical acceleration is equal to gravitational acceleration. Using trigonometry,Angle from the vertical = arctan(25/6 / 9.8) = 23.03 degrees.

(b) Since the car is decelerating, one end of the string attached to the car is pulled back and because of the inertia of the object it is goes on the opposite direction and so the object moves toward the windshield.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
4 0
3 years ago
Pls help with my geometry
Georgia [21]
AB, CA, BC hope this helps
7 0
3 years ago
Read 2 more answers
(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).
gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative \frac{dy}{dx} at that point. By the chain rule,

\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

Differentiate the given parametric equations with respect to t :

x = 4t \implies \dfrac{dx}{dt} = 4

y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}

Then

\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}

We have x=8 and y=2 when t=2, so the slope at the given point is \frac{dy}{dx} = -\frac14.

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}

Alternatively, we can eliminate the parameter and express y explicitly in terms of x :

x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x

Then the slope of the tangent line is

\dfrac{dy}{dx} = -\dfrac{16}{x^2}

At x = 8, the slope is again -\frac{16}{64}=-\frac14, so the normal has slope +4, and so on.

5 0
2 years ago
Need help on this please​
Likurg_2 [28]
The correct scale factor is 3 :)
5 0
3 years ago
I am older than 57 <br> I'm younger than 60<br> I have 9 units
Dahasolnce [82]
Your answer is 59.
HOPE IT HELPS!
7 0
3 years ago
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