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3 years ago

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Two 100kg bumper cars are moving towards eachother in oppisite directions. Car A is moving at 8 m/s and Car B at -10 m/s when th

ey collide head on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision

1 answer:

DerKrebs [107]3 years ago

7 0

Answer:

$-10 m/s$

Explanation:

When two cars collide then the momentum of two cars will remains conserved

Mass of two cars = 100 kg

Speed of car A = 8 m/s

Speed of car B = - 10 m/s

After collision the speed of car B = +8 m/s

By momentum conservation equation

$m1v1i+m2v2i=m1v1f + m2v2f$

$(100)(8)+(100)(-10)=(100v)+(100)(8)\\ v=-10 m/s$

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Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t

ruslelena [56]

Answer:

$x_1= 0.0425m$

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

$m*g = k*x$

With two springs, let, T1 be the tension in each spring, x1 be the extension of each spring. The spring constant of each spring is 2k so:

$T_1 = 2k*x_1$

$2T_1 = m*g=4k x_1$

Solve to x1

$x_1=\frac{m*g}{4k}$

$x_1=\frac{k*x}{4*k}$

$x_1=\frac{x}{4}$

$x_1 = 0.170 / 4$

$x_1= 0.0425m$

7 0

2 years ago

The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r

Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity. Therefore:

marble > manhole cover > basketball > wedding ring

7 0

3 years ago

A body weighing 108N moves with speed of 5m/s in a horizontal

poizon [28]

Answer:

i) 5 $m$ / $s^{2}$ ii) 54 $N$ iii) 54 $N$

Explanation:

i) $a = \frac{v^{2}}{r}$ ⇒ $a$ = 5² ÷ 5 = 5 $m/s^{2}$

ii) $m = \frac{W}{g}$ ⇒ $m$ = 108 ÷ 10 = 10.8 $kg$ , $F = ma$ ⇒ $F =$ 10.8 × 5 = 54 $N$

iii) F1 = F2 = 54 $N$

4 0

2 years ago

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l

Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

$u=+6.3 m/s$ is the initial velocity (positive because it is upward)

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

$v=+6.3 + (-9.8)(1.7)=-10.4 m/s$

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0

3 years ago

Read 2 more answers

How much tension must a rope withstand if it is used to accelerate a 6526 kg car vertically upwards at 8.9 m/s^2?

ExtremeBDS [4]

Answer:

The value is $T = 122036.2 \ N$

Explanation:

From the question we are told that

The mass of the car is $m = 6526 \ kg$

The acceleration is $a= 8.9 \ m/s^2$

Generally the net force applied on the rope is mathematically represented as

$F_{net} = T - W$

Here W is the weight of the car which is evaluated as

$W = m * g$

=> $W = 6526 * 9.8$

=> $W = 63954.8 \ N$

Generally the net force can also be mathematically represented as

$F = m * a$

So

$m * a = T - 63954.8$

=> $6526 * 8.9 = T - 63954.8$

=> $T = 122036.2 \ N$

7 0

2 years ago