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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A wooden block moves at a constant speed on a rough horizontal surface. Draw a free-body diagram
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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

                                              ↓

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Answer:

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Part IV Objects on an incline w/ Tension + Friction
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0.25
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Answers

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

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The given parameters;

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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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Answer:

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Answer:

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Hey mate....

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This is ur answer.....

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Hi there!

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Answer:

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

Explanation:

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Answers

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Frequency means waves passing per second. So 10Hz means 10 waves passing a second

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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.
Determine is the acceleration of the ball.
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:

equation of the distance (0.135m) is:

[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);

also the required acceleration is:

[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.

A block with initial velocity of 3 m/s slides 9 m across a rough horizontal surface before coming to rest. What is the coefficient of kinetic
friction?

Answers

The coefficient of kinetic friction between the block and the horizontal surface is 0.051.

The given parameters;

initial velocity, u = 3 m/sdistance traveled by the block, s = 9 m

The acceleration of the block is calculated as follows;

[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]

The coefficient of kinetic friction between the block and the horizontal surface is calculated as follows;

[tex]\mu_k mg = ma\\\\\mu_k = \frac{a}{g} \\\\\mu_k = \frac{0.5}{9.8} \\\\\mu_k = 0.051[/tex]

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A 455 kg cannon fires a 21 kg cannonball at 26 m/s. If the cannon is on wheels, at what velocity does it move backward?

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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.

total momentum of cannon and cannon ball before = 0 kg m/s - because neither object is moving

total momentum of cannon and cannon ball after collision = 0 kg m/s - because momentum is conserved

Momentum of cannon ball after firing = 4.0 × 120 = 480 kg m/s.

Momentum of cannon after firing = -480 kg m/s (because it recoils in the opposite direction and 480 - 480 = 0 kg m/s, the total momentum after collision).

Rearrange p = m v to find v:

v = p/m

v = 480 divided by 96

v = 5.0 m/s

Note that the forward velocity of the cannon ball was given a positive value. The negative value for the cannon's velocity shows that it moved in the opposite direction.

if the amplitude of a sound wave is tripled, the intensity will:

Answers

Answer:

Increase 9 times

Explanation:

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