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chubhunter [2.5K]

3 years ago

13

A race car is moving with a velocity of 144 kilometers/hour. The driver applies the brakes, and the car comes to a halt in 12.0

seconds. What is the average acceleration of the car during those 12.0 seconds?

1 answer:

anzhelika [568]3 years ago

8 0

-12 k/s/s (the number is right the abbreviation might not be)

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At an amusement park, a swimmer uses a water slide to enter the main pool. a. If the swimmer starts at rest, slides with negligi

adoni [48]

Answer:

a)6.7m/S

b)6.8m/s

Explanation:

Hello ! To solve the point b you must follow the steps below

1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)

2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)

3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)

4)

the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation

$Vfy= \sqrt{Vy^2+2gy}$

where

Vfy= final verticaly speed

Vy=initial verticaly speed=0.59m/S

g=gravity=9.81m/S^2

y=height of slide=2.31m

solving

$Vfy= \sqrt{Vy^2+2gy}\\Vfy= \sqrt{(0.59)^2+2(9.81)(2.31)}=6.77m/s$

The last step is to add the velocity components vectorally at the end of the route with the following equation

$V=\sqrt{Vfy^2+Vx^2} =\sqrt{6.77^2+0.59^2} =6.8m/s$

point A

taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.

$Vfy= \sqrt{Vy^2+2gy}$

vy=0

$Vfy=\sqrt{2gy}$

Vfy= $\sqrt{2(9.81)(2.31)}$ =6.7m/s

8 0

3 years ago

Identify the type of weathering seen in each picture

alukav5142 [94]

Where is the picture
?

5 0

3 years ago

Read 2 more answers

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

3 0

3 years ago

Read 2 more answers

• What value is used to identify each element in the periodic table?

bearhunter [10]

Answer:

protons

Explanation:

5 0

3 years ago

A car of m = 1200. kg collides with a tree while traveling 60.0 mph. The collision occurs over a time period of 0.0500 seconds.

MrRissso [65]

You may know linear momentum is given by P= mass.velocity. Initially car is moving with some velocity so you know initial momentum of the car. Finally it comes to rest i.e final momentum of the car is 0. According to Newton's second law : Force = change in momentum /time. Applying this you'll get answer as 642840N. Hope it helped you. Revert back to me if you have any questions. Please check out the calculation it might be wrong!

5 0

3 years ago

Read 2 more answers