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irakobra [83]
3 years ago
7

What are weather balloons?

Physics
1 answer:
victus00 [196]3 years ago
8 0

Answer:

3 Objects that carry instruments into the stratosphere to measure atmospheric conditions

Explanation:

Weather balloons are objects that carries instruments into the stratosphere to measure atmospheric conditions.

They are very important weather measuring device that helps meteorologists in their fore casts.

These balloons are specially designed for high altitudes. They carry very sensitive devices that are very useful in collecting information about temperature, precipitation, pressure e.t.c

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Which is the force of repulsion between two positively-charged particles?
klio [65]
<span> answer>>>>electric force <<<<by the way i don't like physics but i answer this for you ^-^</span>


<span>This is because centripetal force is just the net force of a circular motion. There are no attractive or repulsive forces here. This is not the case here. </span>
<span>The gravitational force is a force reliant on mass and attraction of the masses. There are attractive forces here, but not really repulsive forces. </span>
<span>The electric force is the only one that would make sense because it has to do with a relationship between charges and includes both repulsive and attractive forces.</span>
4 0
3 years ago
Read 2 more answers
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the ma
Nana76 [90]

Answer:

30.63 m

Explanation:

From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

Maximum height (h) =.?

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

t = 2.5 s

Thus, the time (t) taken to reach the maximum height is 2.5 s

Finally, we shall determine the maximum height reached by the ball as follow:

Time (t) taken to reach the maximum height = 2.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

h = ½gt²

h = ½ × 9.8 × 2.5²

h = 4.9 × 6.25

h = 30.625 ≈ 30.63 m

Therefore, the maximum height reached by the cannon ball is 30.63 m

3 0
3 years ago
The strength of an electromagnet can be altered by
denpristay [2]
The strength of an electromagnet can be altered by increasing the number of coils around the core. The more times the coil is wrapped, the stronger the electromagnet is.

Your answer is: B) Increasing the number of coils around the core 

Have an amazing day and stay hopeful!

3 0
3 years ago
Read 2 more answers
How much work is done if a force of 20N is used to move an object 6 metres? <br><br> pls help
Dmitriy789 [7]

I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.

Work = 20 N * 6 m * 1

Work = 120 Nm

Work = 120 joules

Hope this helps!

8 0
3 years ago
Read 2 more answers
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