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ArbitrLikvidat [17]
3 years ago
5

A carpenter apprentice was killed when he was struck in the head by a nail that was fired from a powder actuated tool. The tool

operator, while attempting to anchor a plywood form in preparation for pouring a concrete wall, fired the gun, causing the nail to pass through the hollow wall. The nail traveled some twenty-seven feet before striking the victim. The tool operator had never received training in the proper use of the tool, and none of the employees in the area were wearing personal protective equipment. What should be recommended to prevent this accident from happening again?
Physics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

Explanation:

The recommendations that need to be urgently addressed by the workshop and it's employees are

1) Training of all employees (particularly the tool operators) on the safe and proper use of all tools used in the course of the carpentry work.

2) Training of all employees in health, safety and environment (HSE). This training will help them appreciate the importance of always wearing personal protective equipment (PPE) when at the workshop or during the course of a work (because if the apprentice was putting on an engineer's helmet, he could have been safe).

3) Employees who fail to adhere strictly to the use of PPE should be sanctioned or even dismissed from the workplace

You might be interested in
A stone is thrown horizontally from a 50m high cliff with an initial speed of 15 meters per second. How far will the stone have
IceJOKER [234]

This is a classic case of 'velocity components.'

Imagine a vector for velocity. Now, consider that this vector could be the hypotenuse of a right triangle, with two other sides going along the x and y-axis. These sides of the triangle will have values, and adding them up using the pythagorean theorem will prove that the sum of their squares equals the square of the original vector.

Well, okay, that's nice and all, you may be saying - but how do we solve the actual question?

Let's apply this thought to the question. This vector can have both an x component and y component (essentially, parts of the vector that travel along the x and y-axis).

Now, what could these components be? We know that the stone is thrown perfectly horizontally, meaning that the x-component is quite literally the velocity.

How about the y-component? Since it's thrown at a perfect horizontal, there isn't really any vertical velocity whatsoever. There's only horizontal velocity.

"Great, fantastic! So, what's the importance of figuring out the horizontal and vertical velocities?"

When a stone is in free fall, it experiences a gravitational acceleration. This acceleration from gravity, though, only affects the vertical velocity. Since gravity is vertical as well, it's essentially impossible for the horizontal velocity to be changed at all.

This means that to solve the horizontal distance, we simply need to find the time it takes for the rock to hit the ground and multiply said time by the horizontal velocity.

Since the vertical velocity is the only thing changed by gravity, we can write out an equation that can solve for the time:

∆Y = v_{i}t + \frac{1}{2}gt^{2}

<u>We know that initial vertical velocity is zero, so:</u>

∆Y = v_{i}t + \frac{1}{2}gt^{2}

∆Y = 0t + \frac{1}{2}gt^{2}

∆Y = \frac{1}{2}gt^{2}

<u>We need to solve for t, so let's isolate the variable. Multiply both sides by 2 to get rid of the fraction:</u>

∆Y * 2 =  \frac{1}{2}gt^{2} * 2

2∆Y = gt^{2}

<u>Divide both sides by g:</u>

(2∆Y)/g = \frac{gt^{2} }{g}

<u>Square root both sides:</u>

\sqrt{(2Y)/g} = \sqrt{t^{2} }

t = \sqrt{(2Y)/g}

<u>Input our values for Y and g (Y is the height of the cliff, and g is gravitational acceleration):</u>

t = \sqrt{(2*50)/9.80}

<u>Solve:</u>

t =  \sqrt{(2*50)/9.80}

t = 3.194 (s)

Whew! That was a lot of steps to find the time! Now that we have the time, we can find the horizontal distance the rock travels:

∆x = v_{i}t + \frac{1}{2}at^{2}

<u>The horizontal velocity has no acceleration (gravity is vertical!), so:</u>

∆x = v_{i}t + \frac{1}{2}*0*t^{2}

∆x = v_{i}t

<u>The horizontal velocity is 15 m/s, and the time is 3.194:</u>

∆x = v_{i}t

∆x = 15 * 3.194

∆x = 47.91 (m)

Since we rounded the time, it makes sense that our final answer's a little bit off to the options. The closest one is option B, which is only 0.6m off, a tiny difference that may have come from the test maker's use of '10 m/s^{2}' as the gravitational acceleration (while we stayed as accurate as possible with 9.80) as well as our rounding of the final time.

Option B, the stone will have travelled 47.85 meters.

If you have any questions on how I got to the answer or if you're still confused on any topic I attempted to explain, just ask in the comments and I'll try to answer it to the best of my ability! Good luck!

- breezyツ

7 0
3 years ago
A nonconducting solid sphere of radius 8.40 cm has a uniform volume charge density. The magnitude of the electric field at 16.8
mina [271]

Answer:

The sphere's volume charge density is 2.58 μC/m³.

Explanation:

Given that,

Radius of sphere R= 8.40 cm

Electric field E= 2.04\times10^{3}\ N/C

Distance r= 16.8 cm

We need to calculate the sphere's volume charge density

Using Gauss's law

\int{\vec{E}\cdot\vec{da}}=\dfrac{Q_{enc}}{\epsilon_{0}}

E\times 4\pi r^2=\dfrac{1}{\epsilon_{0}}\times\dfrac{4}{3}\piR^3\rho

E=\dfrac{\rho R^3}{3\epsilon_{0}r^2}

\rho=\dfrac{3\times E\times\epsilon_{0}r^2}{R^3}

Put the value into the formula

\rho=\dfrac{3\times2.04\times10^{3}\times8.85\times10^{-12}\times(16.8\times10^{-2})^2}{(8.40\times10^{-2})^3}

\rho=2.58\times10^{-6}\ C/m^3

\rho=2.58\ \mu C/m^3

Hence, The sphere's volume charge density is 2.58 μC/m³.

5 0
4 years ago
A 70 kg student is riding on a skateboard at 4m/s. A 60 kg student is also riding a
Fudgin [204]
Answer is C, one student has more mass, therefore more inertia.
6 0
4 years ago
An abstract sculpture consists of a ball (radius R = 76 cm) resting on top of a cube (each side L = 200 cm long). The ball and t
scZoUnD [109]

Answer:

132.9 cm

Explanation:

Data provided in the question:

Radius of the basll = 76 cm = 0.76 m

Side of the box = 200 cm = 2 m

Density of the ball and cube are equal

let the density be 'D'

Now,

Mass of ball, M = Volume × Density

= \frac{4}{3}\pi r^3  × D

= \frac{4}{3}\pi (0.76)^3× D

= 1.838D

Mass of cube, m = L³ × D

= 2³ × D

= 8D

Thus,

center of mass, y = [ My₁ + my₂ ] ÷ [M + m]

here,

y₁ = center of mass of ball with respect to floor

as the center mass of sphere lies at the center of the sphere

= Length of cube + radius of sphere

= 2 + 0.76

= 2.76 m

y₂ = Center of mass of cube = \frac{L}{2}=\frac{2}{2} = 1 m

Thus,

y = [ ( 1.838D × 2.76 ) + (8D × 1 ) ] ÷ [1.838D + 8D]

= 13.07288D ÷ 9.838D

= 1.329 m

or

= 132.9 cm

3 0
3 years ago
The radius of the earth is 6378 km. What is the diameter of the earth in meters?
Nesterboy [21]
To determine the diameter of the earth in metres first multiply the original value by 2.

6378 X 2 = 12 756 km.

Then convert km - m

1 km = 1000 m
12 756 km = ? m

12 756 • 1000 = 12 756 000 = 12 756 000 m or 1.2756 X 10 ^ 7 m

The final solution for the diameter is 1.2756 X 10 ^ 7 m.
7 0
4 years ago
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