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olganol [36]
2 years ago
9

Which should be done in case of a laboratory accident?

Physics
1 answer:
stich3 [128]2 years ago
4 0
Tell your instructor or teacher
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Pros and Cons of the Mercator projection as a scientific model.
Andrew [12]

Answer:

ok i don't know the answer but I send you tomorrow

6 0
2 years ago
An airplane capable of an airspeed of 100 km/hr is 60 km off the coast above the sea. If the wind is blowing from the coast out
Nady [450]

To solve this problem we will apply the concepts related to relative speed. We will obtain it from the deduction made on the aircraft as a speed of the two components that act on it. Through the kinematic equations of motion, we can then calculate the time required.

The airspeed of airplane is 100km/h  while the wind is blowing from the coast out to sea at 40km/h. Wind is blowing from the coast out to sea means that it opposes the airspeed. Therefore, resultant relative speed of airplane is

v_r = 100-40=60km/h

Total distance is 60km then with this net velocity we have that the required time is

v = \frac{x}{t} \rightarrow t = \frac{x}{v}

Where,

x = Displacement

t = Time

v = Velocity

Replacing,

t = \frac{60km}{60km/h} = 1hour

t = 60 minutes

Therefore the time taken by the plane to reach the shore is 60 minutes

6 0
2 years ago
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min
jekas [21]

Answer:

a)  1.301 kg/s

b) 0.001301 m³/s

c) V₁ = 6.505 m/s, V₂ = 1.626 m/s

d) 118.93 kPa

Explanation:

Given:

The number of cans  = 220

The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³

time = 1 minute = 60 seconds

gauge pressure at point 2, P₂ = 152 kPa

b) Thus, the volume flow rate, Q = Volume/ time

Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s

a) mass flow rate = Volume flow rate × density

since it is mostly water, thus density of the drink = 1000 kg/m³

thus,

mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s

c) Given:

Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²

Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²

also,

Q = Area × Velocity

thus, for point 1

0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)

or

V₁ = 6.505 m/s

for point 2

0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)

or

V₂ = 1.626 m/s

d) Applying the Bernoulli's theorem between the points 1 and 2 we have

P_1+\rho gV_1 + \frac{\rho V_1^2}{2}=P_2+\rho gV_2 + \frac{\rho V_2^2}{2}

or

P_1=P_2+\rho\timesg(y_2-y_1)+\frac{\rho}{2}(V_2^2-V_1^2))

on substituting the values in the above equation, we get

P_1=152+1000\times 9.8(1.35)+\frac{1000}{2}(1.626^2-6.505^2))

it is given that point 1 is above point 2 thus, y₂ -y₁ is negative

or

P_1=118.93\ kPa

thus, gauge pressure at point 1 is 118.93 kPa

8 0
3 years ago
The diver on the diving board is 10 meters high and has a mass of 0.050kg, what is his GPE?
sveta [45]

Answer:

<h2>4.9 J</h2>

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 10 × 9.8 × 0.05

We have the final answer as

<h3>4.9 J</h3>

Hope this helps you

6 0
3 years ago
Activity
qwelly [4]

Answer:Eating. Your muscles in your arms and mouth use energy to feed itself. Then your body digest the food which also takes energy.

Sleep. When your tired, you don’t have much energy. It is said that you use more energy while your sleeping. But how do you become energized if you were using even more energy than before?

6 0
2 years ago
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