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Romashka-Z-Leto [24]
2 years ago
15

Help me please (*˘︶˘*).。*♡​

Physics
2 answers:
lisov135 [29]2 years ago
7 0

Answer:

<h3>The angle of contact does not depend upon</h3><h2><em>The inclination of surface in contact</em>.</h2>

Explanation:

Except The inclination of surface in contact, the angle of contact depends upon the temperature, soluble impurity and on cohesive force...

ss7ja [257]2 years ago
5 0

Answer:

your answer is number 4

Explanation:

Hope it helps :)

pls mark brainliets :P

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What is the enthalpy change, ΔH, for this reaction? Show your work.
zubka84 [21]

Answer:

B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.

Explanation:

B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.

3 0
3 years ago
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What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m
Sergio039 [100]

Answer:   
 M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b] 
 Î”R is the difference in density between the gas inside and surrounding the balloon. 
 R[b] is the density of gas inside the baloon.   
 ==================================== 
 Let V be the volume of helium required. 
 Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V   
 U = 1.225gV newtons 
 ---- 
 Weight of Helium = Volume of Helium * Density of Helium * g 
 W[h] = 0.18gV N   
 Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

 
 Weight of 260kg = 2549.7 N 
 Then to lift the whole thing, F > 2549.7 
 So minimal F would be 2549.7 
 ---- 
 1.045gV = 2549.7 
 V = 248.8 m^3   
 Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)   
 =====   
 Let the density of the surroundings be R 
 Then U-W = (1-0.9)RgV = 0.1RgV   
 So 0.1RgV = 2549.7 N 
 V = 2549.7 / 0.1Rg   
 Assuming that R is again 1.255, V = 2071.7 m^3 
 Then mass of hot air required = 230.2 * 0.9R = 2340 kg   
 Notice from this that M = 2549.7/0.9Rg * 0.1R so   
 M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)   
 M[min] = M[basket] * ΔR/R[b]
3 0
3 years ago
A boat sails along the shore. To an observer, the boat appears to move at a speed of 22 m/s, and a man on the boat walking forwa
seropon [69]
The boat is moving at 22 m/s while the man is moving at 23.1 m/s.

That means the man, relative to the boat, is moving at 23.1-22 = 1.1 m/s.

v =d/t, so t = d/v --> t = 3/1.1 = 2.7 s
7 0
3 years ago
A red rubber ball rolls down a hill from rest with an acceleration of 7.8 m/s 2 . How fast is it moving after it has traveled 5
masya89 [10]

Answer:

39 m/s

Explanation:

7 0
3 years ago
An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =
Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V_m = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V_m =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

I(t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

I(t) = V₀√(C/L)

we substitute

I(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

I(t) = 25 × 0.00002

I(t) = 0.0005 A

I(t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0
3 years ago
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