No official standard exists for extinguisher-cylinder colors. that said, a red cylinder is typically associated with which type

through the Doppler effect. The formula for apparent frequency is derived as F apparent = F real x (Vair ± Vobserver) / (Vair ± Vsource). In this scenario, should the observer move towards the source—place a positive sign in the numerator and a negative in the denominator. Since the observer approaches the wall, we apply the formula to derive the necessary speed.

Answer:

Speeds of 1.83 m/s and 6.83 m/s

Explanation:

Based on the law of conservation of momentum,

where m represents mass, is the initial speed before impact, and are the velocities of the impacted object after the collision and of the originally stationary object after the impact.

Thus,

After the collision, the kinetic energy doubles, therefore:

Substituting the initial velocity of 5 m/s provides the equation needed to proceed.

We know that leads to

Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25.

By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

                        (10-5/12) × (21) = 218.75 meters.

At time , the ball's horizontal and vertical velocities can be represented as

However, since the ball is thrown horizontally, we have . The horizontal and vertical positions at time are

The ball travels a distance of 22 m horizontally from the throw point, thus

With this, we determine that the time for the ball to reach the ground is

When it touches down, and

1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields


3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is

, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:

, carrying a negative sign to denote a downward direction.