ASAP
A train is travelling at 100 m/s when it receives an alert: “Brake! 600 meters ahead, the rails are broken”. The highest acceleration that the train can produce is 10 m/s^2. If the train slows down with a constant acceleration of 10 m/s^2
How long will it take for the train to stop?
How many meters will it travel before it stops?
Botcookiemaster please don't use a file just give me the answer

Answers

Answer 1

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Let's consider the given terms ~

Initial velocity (u) = 100 m/s

Acceleration (a) = - 10 m/s²

(Acceleration is negative bbecause train slows down)

Final velocity (v) = 0 m/s

(final velocity is 0, because train stops)

Now, let's find the time taken (t) by using the first equation of motion ~

[tex]v = u + at[/tex]

[tex]0 = 100 + ( - 10 \times t)[/tex]

[tex]0 = 100 - 10t[/tex]

[tex]10t = 100[/tex]

[tex]t = 100 \div 10[/tex]

[tex]t = 10[/tex]

The train will take 10 secs to stop,

Now, let's find how much distance (s) it will cover before it stops using second equation of motion ~

[tex]s = ut + \dfrac{1}{2} at {}^{2} [/tex]

[tex]s = (100 \times 10) + ( \dfrac{1}{ 2} \times - 10 \times 10 \times 10)[/tex]

[tex]s = 1000 + ( - 500)[/tex]

[tex]s = 1000 - 500[/tex]

[tex]500 \: \: m[/tex]

It will cover 500 meters before coming to rest ~


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Answer:

A cannon ball of mass 4.0 kg is fired from a stationary 96 kg cannon at 120 m/s. Calculate the velocity of the cannon immediately after firing.

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Answers

Hi there!

To solve, we can use the following equation:

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Answer:

[tex]\boxed {\boxed {\sf 27 \ m/s}}[/tex]

Explanation:

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[tex]v^2 = u^2 + 2(-a)s\\\\v^2 = u^2 - 2as\\\\a = \frac{u^2 - v^2}{2s} \\\\a = \frac{3^2 - 0^2}{2\times 9} \\\\a =0.5 \ m/s^2[/tex]

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Alejandro Kirk is the catcher for the Blue Jays’ baseball team. He exerts a forward force on the 0.145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m.
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Determine the force applied by Alejandro.

Answers

Answer:

a= -5404.6 [m/s²]; F=785.75 [N].

Explanation:

1. Determine is the acceleration of the ball:

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[tex]S=-\frac{at^2}{2} +v_0t,[/tex]  where S=0.135[m]; a - required acceleration; t - elapsed time; v₀ - initial velocity (38.2 m/s);

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[tex]a=\frac{v-v_0}{t}, where[/tex] V - the end velocity (0 m/s), t - elapsed time, ₀ - initial velocity (38.2 m/s).

Using the equations of reqruired acceleration and the distance it is possible to make up and solve the next system:

[tex]\left \{ {{a=-\frac{38.2}{t} } \atop {S=-\frac{at^2}{2}+38.2t }} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {0.5at^2-38.2t=0.135}} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {19.1t-38.2t=0.135}} \right. \ => \ \left \{ {{a=-5404.6} \atop {t=0.007}} \right.[/tex]

finally, a≈-5404.6 [m/s²].

2. Determine the force applied by Alejandro.

the energy is:

[tex]E=\frac{mv^2}{2}; \ or E=FS, where \ m-the \ mass; \ v-velocity; \ F-required \ force; S-distance;[/tex]

According to these two equations, the required force is:

[tex]F=\frac{E}{S}=\frac{mv^2}{2S};[/tex]

F=0.145*38.8²/2/0.135≈785.75 [N].

note, the suggested way is not the shortest one and not the only one.

P.S. if it is possible, check the arithmetic operations and the provided answers in other sources.

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Answers

The weight of the block and normal reaction are equal in magnitude but opposite in direction. Also, the frictional force and applied force are equal in magnitude but opposite in direction.

The forces acting on the wooden block moving at a constant speed include the following;

weight of block acting downwards, Wnormal reaction on the block acting upwards, Napplied force on the block acting towards positive x-direction, Ffrictional force acting towards negative x-direction, [tex]F_f[/tex]

The free-body diagram is presented below;

                                         

                                              N

                                              ↑

                               [tex]F_f[/tex]      ←  ⊕  →   F

                                              ↓

                                              W

The block is moving at a constant speed, we will have the following;

[tex]W = N \\\\F = F_f[/tex]

Thus, we can conclude that the weight of the block and normal reaction are equal in magnitude but opposite in direction. Also, the frictional force and applied force are equal in magnitude but opposite in direction.

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Part IV Objects on an incline w/ Tension + Friction
1. A small 63 kg sleigh is pulled by a rein attached to horse up a 15'angle hill to the
horizontal. The tension of the rein is 510 N. The coefficient of kinetic friction is
0.25
a. What is the normal force that the sleigh exerts on the hill?
b. What are the magnitude and direction of the sleigh's acceleration?

Answers

(a) The normal force on the sleigh is 596.36 N.

(b) The magnitude and direction of acceleration of the sleigh is 3.2 m/s² upwards.

The given parameters;

mass of the sleigh, m = 63 kginclination of the hill, θ = 15⁰tension on the rein, F = 510 Ncoefficient of friction, μ = 0.25

The normal force on the sleigh is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 63 \times 9.8 \times cos(15)\\\\F_n = 596.36 \ N[/tex]

The magnitude and direction of acceleration of the sleigh is calculated as follows;

[tex]\Sigma F= ma\\\\F - mgsin(\theta) - F_f = ma\\\\F - mgsin(\theta) - \mu F_n = ma\\\\510\ - \ 63 \times 9.8 \times sin15 \ -\ 0.25\times 596.36 = 63a\\\\201 .11 = 63a\\\\a = \frac{201.11}{63} \\\\a = 3.2 \ m/s^2 \ upwards[/tex]

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What do scientists call a substance that contains atoms from two or more different elements that are chemically bonded?

Atom
Compound
Element
Molecule

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Answer is b compound is a substance that contains two or more elements chemically combined in a fixed proportion. The elements carbon and hydrogen combine to form many different compounds.

Find the acceleration if the force is 10 N and the mass is 5000 g.

Answers

Answer:

If the force is 10N and mass is 5000G, then a=2m/s2

Explanation:

F=10N

m=5000

g=5kg

a=10N/ 5kg

and finally; a= 2m/s2

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