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Viefleur [7K]
2 years ago
8

Two cars A and B are 100m apart moving towards each other with

Physics
1 answer:
maxonik [38]2 years ago
3 0

Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

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slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

W = F_{total} .d

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W  = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} .d =\frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

F_{sprinter} = F_{total} + F_{wind}  = 39.7 + 30 = 69.68 N

7 0
3 years ago
Read 2 more answers
Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S <br>​
Tems11 [23]

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}

6 0
3 years ago
2 When a cube of hot metal is placed in a beaker of cold water, the temperature of the water -
jek_recluse [69]
The answer to this is B
4 0
3 years ago
you give a shopping cart a job down the aisle the cart is full of groceries and has a mass of 18 kilograms the cart accelerates
mezya [45]

                 Force = (mass) x (acceleration)

                 Force = (18 kg) x (3 m/s²) = 54 newtons

As long as you continue pushing the cart with 54 newtons of force,
it will accelerate at 3 m/s². 

At the instant you release it, or keep your hands on it but stop pushing,
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6 0
2 years ago
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How will the motion of a falling whirligig compare to that of a falling paper ball?
kicyunya [14]

The motion of a falling whirligig is different to that of a falling paper ball due to spinning.

<h3>Type of motion performed by whirligig and falling paper ball </h3>

The motion of a falling whirligig is different from the motion of a falling paper ball because the paper ball falls on the ground without spinning while on the other hand, the whirligig falls on the ground along with spinning.

The falling whirligig performs two motion i.e. one is falling on the ground and the other is spinning during motion whereas paper ball performs one motion i.e. motion in the air towards the ground so we can conclude that the motion of a falling whirligig is different than of a falling paper ball.

Learn more about motion here: brainly.com/question/453639

4 0
2 years ago
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