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3 years ago

Which type of foot PPE must you wear if electrical shock is ever a possibility with your job?

1 answer:

7 0

The appropriate PPE(personal protective equipment) to be worn for foot protection against electric shock is boots retrofitted with rubber soles. Rubber does not conduct electricity and so will safeguard against electric shocks.

Other personal protective equipment used in a lab or industry can include safety glasses, rubber gloves, protective clothing and masks and respirators. PPE goes hand in hand with personal behaviour, personal grooming and housekeeping.

You might be interested in

Which statement correctly describes a characteristic that a scientific measuring tool should have

Reil [10]

Explanation:

To be accurate, it must be able to make measurements that are close to the actual value.

6 0

3 years ago

A raft with the area A , thickness= h and the mass 600 kg, Floats in still water with 7 cm

elena55 [62]

In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.

ρwAd1g = Mg

ρwAd2g = (M + m) g

d2∕d1 = (M + m)/g

m = [(d2∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg

This means that Bubba’s mass is 120 kg.

7 0

3 years ago

4) A satellite, mass m, is in circular orbit (radius r) around the earth, which has mass ME and radius Re. The value of r is lar

defon

<h2>Answers:</h2>

(a) The kinetic energy of a body is that energy it possesses due to its movement and is defined as:

$K=\frac{1}{2}m{V}{2}$ (1)

Where $m$ is the mass of the body and $V$ its velocity.

In this specific case of the satellite, its kinetic energy $K_m$ taking into account its mass $m$ is:

$K_{m}=\frac{1}{2}m{V}^{2}$ (2)

On the other hand, the velocity of a satellite describing a circular orbit is constant and defined by the following expression:

$V=\sqrt{G\frac{ME}{r}}$ (3)

Where $G$ is the gravity constant, $ME$ the mass of the Earth and $r$ the radius of the orbit (measured from the center of the Earth to the satellite).

Now, if we substitute the value of $V$ from equation (3) on equation (2), we will have the final expression of the kinetic energy of this satellite:

$K_{m}=\frac{1}{2}m{\sqrt{G\frac{ME}{r}}}^{2}$ (4)

Finally:

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5) >>>>This is the kinetic energy of the satellite

(b) According to Kepler’s 2nd Law applied in the case of a circular orbit, its Period $T$ is defined as:

$T=2\pi\sqrt{\frac{r^{3}}{\mu}}$ (6)

Where $\mu$ is a constant and is equal to $GME$ . So, this equation in these terms is written as:

$T=2\pi\sqrt{\frac{r^{3}}{GME}}$ (7)

As we can see, the Period of the orbit does not depend on the mass of the satellite $m$ , it depends on the mass of the greater body (the Earth in this case) $ME$ , the radius of the orbit and the gravity constant.

(c) The gravitational force described by the law of gravity is a central force and therefore is a conservative force. This means:

1. The work performed by a gravitational force to move a body from a position A to a position B only depends on these positions and not on the path followed to get from A to B.

2. When the path that the body follows between A and B is a closed path or cycle (as this case with a circular orbit), the gravitational work is null or zero.

<h2>This is because the gravity force that maintains an object in circular motion is a centripetal force, that is, it always acts perpendicular to the movement. </h2>

Then, in the case of the satellite orbiting the Earth in a circular orbit, its movement will always be perpendicular to the gravity force that attracts it to the planet, at each point of its path.

(d) The total Mechanical Energy $E$ of a body is the sum of its Kinetic Energy $K$ and its Potential Energy $P$ :

$E=K+P$ (8)

But in this specific case of the circular orbit, its kinetic energy will be expresses as calculated in the first answer (equation 5):

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5)

And its potential energy due to the Earth gravitational field as:

$P_{m}=-G\frac{mME}{r}$ (9)

This energy is negative by definition.

So, the total mechanical energy of the orbit, also called the Orbital Energy is:

$E=\frac{1}{2}Gm\frac{ME}{r}+(- G\frac{mME}{r})$ (10)

Solving equation (10) we finally have the Orbital Energy:

$E=-\frac{1}{2}mME\frac{G}{r}$ (11)

At this point, it is necessary to clarify that a satellite (or any other celestial body) orbiting another massive body, can describe one of these types of orbits depending on its Orbital Total Mechanical Energy $E$ :

-When $E=0$ :

We are talking about an open orbit in which the satellite escapes from the attraction of the planet's gravitational field. The shape of its trajectory is a parabola, fulfilling the following condition:

$K_{m}=-P_{m}$

Such is the case of some comets in the solar system.

-When $E>0$ :

We are also talking about open orbits, which are hyperbolic, being $K_{m}>P_{m}$

We are talking about closed orbits, that is, the satellite will always be "linked" to the gravitational field of the planet and will describe an orbit that periodically repeats with a shape determined by the relationship between its kinetic and potential energy, as follows:

-Elliptical orbit: Although $E$ is constant, $K_m$ and $P_m$ are changing along the trajectory .

-Circular orbit: When at all times both the kinetic energy $K_m$ and the potential $P_m$ remain constant, resulting in a total mechanical energy $E$ as the one obtained in this exercise. This means that the speed is constant too and is the explanation of why this Energy has a negative sign.

3 0

4 years ago

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

Mamont248 [21]

Answer:

74.86°C

Explanation:

P₂ = Vapour pressure of water at sea level = 760 mmHg

P₁ = Pressure at base camp = 296 mmHg

T₂ = Temperature of water = 373 K

ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol

R = Gas constant = 8.314 J/mol K

From Claussius Clapeyron equation

$ln\frac{P_2}{P_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\\Rightarrow ln\frac{760}{296}=\frac{40700}{8.314}\left(\frac{1}{T_1}-\frac{1}{373}\right)\\\Rightarrow ln\frac{760}{296}\times \frac{8.314}{40700}+\frac{1}{373}=\frac{1}{T_1}\\\Rightarrow 0.0028735=\frac{1}{T_1}\\\Rightarrow T_1=347.996\ K$

T₁ = 347.996 K = 74.86°C

∴Water will boil at 74.86°C

4 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$