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2 years ago

What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?

Physics

2 answers:

Elodia [21]2 years ago

8 0

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

AveGali [126]2 years ago

6 0

Equation you need to use is a=Vf-Vi/t
Vi=initial velocity
Vf= Final Velocity
Write out the variables and their numbers
Vf=75 mi/hr Vi=15 mi/hr t=4 seconds
Plug it all in
75-15/4
60/4
a=15 mi/hr^2

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

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3 years ago

HeLp aSAp!!! If the speed and distance of an object are given, and I need to find the time, I will...

Helen [10]

Answer:

B - Divide Speed and Distance

Explanation:

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miv72 [106K]

Answer:

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3 years ago

A crash test dummy with a mass of 55kg is placed into a car traveling at 40m/s. During the first trial the dummy does not wear a

s2008m [1.1K]

a. The force experienced by the dummy with a seatbelt is 11,000 Newton.

b. The force experienced by the dummy without a seatbelt is 4,400 Newton.

<u>Given the following data:</u>

Mass = 55 kg

Initial velocity = 0 m/s

Final velocity = 40 m/s

Time A = 0.2 seconds

Time B = 0.5 seconds

a. To calculate the force experienced by the dummy with a seatbelt:

<h3>Newton's Second Law of Motion</h3>

In order to solve for the force acting on the dummy, we would apply Newton's Second Law of Motion.

<u>Note:</u> The acceleration of an object is equal to the rate of change in velocity with respect to time.

Mathematically, Newton's Second Law of Motion is given by this formula;

$F = \frac{M(v\;-\;u)}{t}$

Substituting the given parameters into the formula, we have;

$F = \frac{55(40\;-\;0)}{0.2}\\\\F = \frac{55\times 40}{0.2}\\\\F = \frac{2200}{0.2}$

Force = 11,000 Newton.

b. To calculate the force experienced by the dummy without a seatbelt:

$F = \frac{55(40\;-\;0)}{0.5}\\\\F = \frac{55\times 40}{0.5}\\\\F = \frac{2200}{0.5}$

Force = 4,400 Newton.

Read more on force here: brainly.com/question/1121817

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