What do parallel circuits and series circuits have in common

Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when

nignag [31]

<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

v₁ is the final speed of car X.

$m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s$

So, car X will move with a velocity of -12 m/s.

3 0

2 years ago

A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c

satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

Given the data in the question;

First course : $a_{x}$ = 0.75 m/s², $d_{x}$ = 20 m, $u_{x}$ = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

$v_{x}$ ² = (10)² + (2 × 0.75 × 20)

$v_{x}$ ² = 100 + 30

$v_{x}$ ² = 130

$v_{x}$ = √130

$v_{x}$ = 11.4 m/s

for the Second Course:

$u_{y}$ = 11.4 m/s, $a_{y}$ = -1.15 m/s², $v_{y}$ = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) × $d_{y}$ )

0 = 129.96 - 2.3 $d_{y}$

2.3 $d_{y}$ = 129.96

$d_{y}$ = 129.96 / 2.3

$d_{y}$ = 56.5 m

so;

|d| = √( $d_{x}$ ² + $d_{y}$ ² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0

2 years ago

The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3

viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

$x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;$

$\Rightarrow mg\sin{\theta} = F$

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at $\theta.$ Solving for the angle, we get