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4 years ago

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Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. car 1 suddenly starts to brake

with constant acceleration and stops in 10 m. at the instant car 1 comes to a stop, car 2 begins to brake with the same acceleration. it comes to a halt just as it reaches the back of car 1. what was the separation between the cars before they starting braking?

1 answer:

Ganezh [65]4 years ago

4 0

<span> Maths delivers! Braking distance ... If the </span>car<span> is initially travelling at u</span>m<span>/s, then the stopping distance d </span>m<span> ... the </span>speed<span> of the </span>car<span> at the </span>instant<span> the </span>brakes<span> are applied. ... An object with </span>constant acceleration<span> travels the </span>same<span> distance as it would ... We </span>start<span> with the second equation of motion:.</span>

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A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away

alekssr [168]

Answer:

31.2 m/s

Explanation:

$f_{app}$ = Frequency of approach = 480 Hz

$f_{aw}$ = Frequency of going away = 400 Hz

$V$ = Speed of sound in air = 343 m/s

$v$ = Speed of truck

Frequency of approach is given as

$f_{app} = \frac{Vf}{V - v}$ eq-1

Frequency of moving awayy is given as

$f_{aw} = \frac{Vf}{V + v}$ eq-2

Dividing eq-1 by eq-2

$\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}$

$\frac{480}{400} = \frac{343 + v}{343 - v}$

$v$ = 31.2 m/s

7 0

3 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0

3 years ago

Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr

Illusion [34]

Answer:

$F_{sprinter}=110.4N$

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

$W=F_{total}.d$

So

$W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N$

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

So

$F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N$

7 0

3 years ago

Read 2 more answers

If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that

alexandr402 [8]

Answer:

$g'=63.74\ m/s^2$

Explanation:

It is given that,

Weight of the person on Earth, W = 818 N

Weight of a person is given by the following formula as :

$W=mg$

g is the acceleration due to gravity on earth

$m=\dfrac{W}{g}$

$m=\dfrac{818\ N}{9.8\ m/s^2}$

m = 83.46 kg

The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

$g'=\dfrac{W'}{m}$

$g'=\dfrac{5320\ N}{83.46\ kg}$

$g'=63.74\ m/s^2$

So, the acceleration due to gravity on that planet is $63.74\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

What voltage battery would you need to send 2.5A of current through a light bulb of resistance 3.6 ohm

nadezda [96]

<h3><u>Given</u> :</h3>

Current flow light bulb = 2.5 $\sf{A}$

Resistance of light bulb = 3.6Ω

<h3><u>To Find </u>:</h3>

We have to find voltage of battery $.$

<h3><u>Solution</u> :</h3>

➠ As per ohm's law, current flow through a conductor is directly proportional to the applied potential difference.

➝ V ∝ I

➝ <u>V = I × R</u>

Where, R is the resistance of conductor.

⇒ V = I × R

⇒ V = 2.5 × 3.6

⇒ <u>V = 9 volt</u>

8 0

3 years ago