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3 years ago

8

Never start a job without knowing the _and_ of the chemicals you are working with a) properties and hazards b) SDS a

nd contents c) stability and flash point d) boiling point and flammable limits

1 answer:

charle [14.2K]3 years ago

6 0

<u>Answer</u>

a) properties and hazards

<u>Explanation</u>

Having information on properties of a chemical helps you know the reactivity of that chemicals.

Most chemicals are very reactive. So you have to be away of their hazards to avoid accidents.

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A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of

Vikentia [17]

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

$\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)$

$mv^2 = mv_t^2 + 2mgR(1 + cos \theta)$

we know that,

$N - mgcos \theta = \dfrac{mv^2}{R}$

$N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}$

$N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)$

$N = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)$

$N = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)$

N = 26.59 N

3 0

3 years ago

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

4 0

3 years ago

Work is done on a locked door that remains closed while you try to pull it open. True False.

dsp73

Answer:False

Explanation:

Work is being done on a body when it causes displacement of body on the application of force

$Work\ done=Force\times displacement$

When we pull the door by a force it causes zero displacements of the door. So we can say that work done on it is zero.

Thus the above-given statement is false

8 0

3 years ago

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is $Q = 5.866 * 10^9 J$

The power is $P = 5866\ MW$

Explanation:

From the question we are told that

Mass of the water per second is $m = 1917 \ kg$

The initial temperature of the water is $T_i = 35^oC$

The boiling point of water is $T_b = 100^oC$

The final temperature $T_f = 450^oC$

The latent heat of vapourization of water is $c__{L}} = 2256*10^3 J/kg$

The specific heat of water $c_w = 4184 J/kg^oC$

The specific heat of stem is $C_s =1520 \ J/kg ^oC$

Generally the heat needed each second is mathematically represented as

$Q = m[c_w (T_i - T_b) + m* c__{L}} + m* c__{S}} (T_f - T_b)]$

Then substituting the value

$Q = m[c_w [T_i - T_b] + c__{L}} + C__{S}} [T_f - T_b]]$

$Q = 1917 [(4184) [100 - 35] + [2256 * 10^3] +[1520] [450 - 100]]$

$Q = 1917 * [3.05996 * 10^6]$

$Q = 5.866 * 10^9 J$

The power required is mathematically represented as

$P = \frac{Q}{t}$

From the question $t = 1\ s$

So

$P = \frac{5.866 *10^9}{1}$

$P = 5866*10^6 \ W$

$P = 5866\ MW$

6 0

3 years ago

What are some of the similar features of position vs time and velocity vs time

qaws [65]

I agree with the other comment

6 0

2 years ago