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zysi [14]
3 years ago
15

A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?

Physics
1 answer:
lapo4ka [179]3 years ago
5 0
<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

V_{f}^{2}-V_{o}^{2}=2.a.d   (1)

Where:

V_{f} is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

V_{o} is the Initial Velocity, the value we want to find

a is the constant acceleration of the car (the negative sign means the car is decelerating)

d is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find V_{o}:

0-V_{o}^{2}=2(-5m/{s}^{2})(15m)    (2)

-V_{o}^{2}=-150{m}^{2}/{s}^{2}    (3)

Multiplying by -1 on both sides of the equation:

V_{o}^{2}=150{m}^{2}/{s}^{2}    (4)

V_{o}=\sqrt{150{m}^{2}/{s}^{2}}    (5)

Finally:

V_{o}=12.24m/s >>>This is the Initial velocity of the car

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a 13-gram bullet, moving at 270 m/s, penetrates a 2 kg block of wood and emerges at a speed of 130 m/s. if teh block sits one a
saveliy_v [14]

Answer:

1.52m/s

Explanation:

Using the law of conservation of momentum

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6 0
3 years ago
A 20-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 26 n. starting from rest, the sle
Vesna [10]
Good morning.

We calculate the acceleration with the <em>Torricelli equation</em>:

\star\  \boxed{\mathsf{V^2 = V_0^2 + 2a\Delta S}}

We see that:

\begin{cases}\mathsf{V_0 = 0}\\\mathsf{V = 2.3 \ m/s}\\\mathsf{\Delta S = 8.9 \ m}\end{cases}

Now:

\mathsf{(2.3)^2 = 0^2 + 2a\cdot 8.9}\\ \\ \mathsf{5.29 = 17.8a}\\ \\ \bold{\mathsf{a = 0.3 \ m/s^2}}


Now we can calculate the resultant force that makes that acceleration of 0.3 m/s² with the 2nd Law of Newton:

\mathsf{F_r = m\cdot a}\\ \\ \mathsf{F_r = 20\cdot 0.3}\\ \\ \mathsf{F_r = 6 \ N}


We have a force of 26 N → and a friction force F ←. Adding those vectors, he have  a force 6 N →. Therefore:

26 - F = 6

F = 20 N


We have a friction force of 20 N. We calculate the kinect coefficient with the formula:

\star \ \boxed{\mathsf{F = \mu_k N}}


Since we are in a horizontal plane, we hava that N = P = mg = 200 N

Therefore:

\mathsf{20 = \mu_k 200}\\ \\ \boxed{\boxed{\mathsf{\mu_k = 0.1}}}
5 0
4 years ago
While on stage introducing the iPhone, Steve jobs walked 3 meters to the right, 9 meters to the left, 7 meters to the right, and
Y_Kistochka [10]

Answer:

Hey

The distance is easy, just add 3+9+7+9.

Answer <em>28m</em>

Distance is not to be confused with displacement (the total distance from your starting point).

<em>Wbob1314</em>

7 0
3 years ago
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