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3 years ago

What information in the MSDS would be most important to know at the end of an experiment?

1 answer:

8 0

The most important information in the MSDS that is useful at the end of an experiment is how to manage or dispose of the waste materials of the experiment. This is important especially if the materials used are toxic. They cannot just be disposed in the sink or the trash bin. They must be disposed in a waste bottle or other methods.

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A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45

vladimir2022 [97]

<span>31.3 m/s Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know. Distance the water balloon travels at velocity v for time t d = vt Total time required for the entire trip is double since the balloon goes up, then goes down t = 2v/a Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2 t = 2v/9.8 100 = vt Substitute 2v/9.8 for t in the 2nd formula 100 = v(2v/9.8) Solve for v. 100 = v(2v/9.8) 100 = 2v^2/9.8 980. = 2v^2 490 = v^2 22.13594 = v So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that 2*22.13594 / 9.8 = 4.51754 So it will take 4.51754 second for the balloon to hit the ground after being launched. 4.51754 * 22.13594 = 100 And during that time it will travel 100 meters horizontally. But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>

3 0

2 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0

2 years ago

Which is not a device for reproducing sound?

Alona [7]

Phonograph. Hope that helps

7 0

2 years ago

a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi

alexandr402 [8]

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, $u=60.0\ mm/s$

Angle of projection is, $\theta=30.0\°$

Time taken to land on the hill is, $t=4\ s$

In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

$u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s$

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

6 0

2 years ago

Calculate the recoil speed of a 1.4 kg rifle shooting 0.006 kg bullets with muzzle speed of 800 m/a.3.43 m/s

Angelina_Jolie [31]

Answer:

a) 3.43 m/s

Explanation:

Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the bullet is shot is zero, because they are both at rest, so:

$p_i = 0$