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3 years ago

6

You are the driver of the car in the photos above. You Are traveling at 30 mph when suddenly the car goes from its position in t

he first photo to the position in the second photo. What is happening

1 answer:

Marrrta [24]3 years ago

5 0

Answer:

the car uses teleportation, to zip to one side of the photo, to the other

Explanation:

You might be interested in

A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg

Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s. Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0

3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

3 years ago

Where would you find the lowest-density seawater?

PolarNik [594]

In rivers that are close to sea

4 0

4 years ago

Imagine that you pushed a box, applying a force of 60 newtons, over a distance of 4 meters. How much would you have done?

grin007 [14]

You would have done 240 joules of work on the box.

Work = (force) x (distance)

= (60 newtons) x (4 meters)

= 240 joules .

5 0

3 years ago

A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mas

Sergio039 [100]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The distance traveled in horizontal direction is $D = 1.38 m$

Explanation:

From the question we are told that

The length of the string is $L = 1.6 \ m$

The mass of the ball is $m = 200 g = \frac{200}{1000} = 0.2 \ kg$

The height of ball is $h = 1.5 \ m$

Generally the work energy theorem can be mathematically represented as

$PE = KE$

Where PE is the loss in potential energy which is mathematically represented as

$PE =mgh$

Where h is the difference height of ball at A and at B which is mathematically represented as

$h = y_A - y_B$

So $PE =mg(y_A - y_B)$

While KE is the gain in kinetic energy which is mathematically represented as

$KE = \frac{1}{2 } (v_b ^2 - 0)$

Where $v_b$ is the velocity of the of the ball

Therefore we have from above that

$PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)$

Making $v_b$ the subject we have

$v_b = \sqrt{2g (y_A - y_B)}$

substituting values

$v_b = \sqrt{2g (1.5 - 0.40)}$

$v_b = 4.6 \ m/s$

Considering velocity of the ball when it hits the floor in terms of its vertical and horizontal component we have

$v_x = 4.6 m/s \ while \ v_y = 0 m/s$

The time taken to travel vertically from the point the ball broke loose can be obtained using the equation of motion

$s = v_y t - \frac{1}{2} g t^2$

Where s is distance traveled vertically which given in the diagram as $s = -0.4$

The negative sign is because it is moving downward

Substituting values

$-0.4 = 0 -\frac{1}{2} * 9.8 * t^2$

solving for t we have

$t = 0.3 \ sec$

Now the distance traveled on the horizontal is mathematically evaluated as

$D = v_b * t$

$D = 4.6 * 0.3$

$D = 1.38 m$

8 0

3 years ago