The diagram shows monochromatic light passing through two openings.

A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

$\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}$

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0

1 year ago

Sean climbs a tower that is 71.3 m high to make a jump with a parachute. The mass of Sean plus the parachute is 81.4 kg. If U =

myrzilka [38]

Answer:

U = 56877.4 J

Explanation:

The potential energy of a body is that which it possesses because it is located at a certain height above the surface of the earth and can be calculated using the following formula:

U = mgh Formula (1)

Where:

U is the potential energy in Joules (J)

m is the mass of the body in kilograms (kg)

g is the acceleration due to gravity (m/s²)

h is the height at which the body is found from the surface of the earth in meters (m)

Data

m= 81.4 kg

g= 9.8 m/s²

h = 71.3 m

Potential energy of Sean and the parachute at the top of the tower

We replace data in the formula (1)

U = m*g*h

U = (81.4 kg)*(9.8 m/s²)*(71.3 m)

U = 56877.4 N*m

U = 56877.4 J

3 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago

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

I NEED HELP ASAP!! I NEED IT BEFORE TOMORROW!! D: 10pts+5pts!

Wewaii [24]

9. Compounds can form from two nonmetals by sharing their electrons in a
C) covalent

11. An atom that has an excess positive or negative electrical charge caused by the loss or addition of an electron is called a(n) ______.
B) ion

5 is either A or C

4 0

3 years ago