A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially

Answers

Answer 1

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s


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I think this is the solution:

1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
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11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4

Question 2 of 25

If JKLM is a parallelogram, what is the length of LM?

K

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8

M

O A. 10

O B. 18

O C. 8

D. 26

SUBMIT

Answers

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Explanation:

Answer:

8

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Answers

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Explanation:

A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is noted to be 15 m/s downward and 12 m/s upward. What is the contact time for the ball if the force applied on the ball from the ground is equal to 10 N?

Answers

Answer:

13.5

Explanation:

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Final Velocity: 12

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We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

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Plugging them in into desmos gives 13.5 for time.

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Thus, the tension in the cable between the blocks, T₂, is [tex]\frac{2T_i}{7}[/tex].

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Answer:

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An object is released from height of 17m.
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[tex]\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12 gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}[/tex]

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PLSSS HELP WILL MARK AS BRIANLIEST
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As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...

Answers

Answer:

As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...

Explanation:

As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...

[tex] \boxed{ \sf \: better \: luck \: at \: your \: class}[/tex]

A uniform solid sphere revolves around a vertical support on a frictionless bearing.

What is Solid sphere?

A solid disk is covered with a light string that circles the sphere's equator, over which is suspended a mass. The mass M of the hanging mass, disk, and sphere is the same.

The radius R of the sphere and disk is the same. If the string does not slip on the sphere or disk, determine the acceleration of the hanging mass. If necessary, explain your response in terms of M, R, and g.

A uniform solid sphere revolves around a vertical support on a frictionless bearing, as seen in the illustration below. Around the sphere's equator, a thin cable travels over a solid surface.

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Three particles are placed in the xy plane. A 30-g particle is located at (3, 4) m, and a 40-g particle is located at (-2, -2) m. Where must a 20-g particle be placed so that the center of mass of the three-particle system is at the origin?

Answers

Answer:

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At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, how much time does it take the puck to come to rest?

Answers

Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.

At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is  4.51 sec.

What is speed?

The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.

Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,

force = ma = μmg, putting the value,

a = - 1.175 m/sec²

now using equation of motion

v = u + at

t = 4.51 sec

So, time taken by the puck to come to rest is 4.51 sec.

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