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Alexxx [7]
3 years ago
14

PLEASE HELP ASAP!!!!!!!!!!!!

Physics
1 answer:
AleksandrR [38]3 years ago
3 0

a = acceleration of swati during free fall = - 9.8 meter/second²

v₀ = initial velocity of swati = 0 meter/second

t = time of travel = 120 seconds

d = displacement of swati during free fall

Using the kinematics equation

d = v₀ t + (0.5) a t²

inserting the values

d = (0) (120) + (0.5) (- 9.8) (120)²

d = - 70560 m

hence distance travelled by swati is 70560 m

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If a material were being designed to demonstrate no thermal expansions, how would the energy plot look like?
bixtya [17]

The characteristics of thermal expansion allow finding that the response for a material without thermal expansion is

  • The length variation is zero
  • In the graph the line is horizontal so there is no change in length with temperature

Thermal expansion is the macroscopic sum of the changes in the length of the bonds when the energy (temperature) changes, it can be written

              ΔL = α L₀ ΔT

Where ΔL is the change in length, α the coefficient of linear expansion, L₀ the initial length and ΔT the change in body temperature

In this case, a material is being designed that the thermal expansion is very small, for this the material must be made up of several compounds where some of them present a contraction with temperature, some examples: water at low temperature, liquefied gases , ceramic tile, quartz, etc.

The thermal expansion measurement processes control the body temperature and measure the change in length, in this case the change in length must be zero, in the attachment we can see a graph of a composite material with these characteristics, an example of this type of material is Invar an alloy of nickel and iron α = 3.7 10⁻⁶ ºC⁻¹

In conclusion, using the characteristics of thermal expansion we can find that the response of material without thermal expansion is

  • The length variation is zero
  • In the graph the line is horizontal so there is no change in length with temperature

Learn more here:  brainly.com/question/18717902

5 0
3 years ago
On a cold, 5°F day, the nosewheel tire of a Learjet, having an internal volume of 1000. in3 is inflated to 30 psi (i.e., 30 lb/i
Lunna [17]

Answer:

The weight in (pounds ) is m  = 0.14994 \  lb

The specific volume is V_s = 0.2403 \  m^3 /kg

Explanation:

From the question we are told that

The temperature is T  =  5^o F  =  (5^oF - 32) *  \frac{5}{9}  =  -15 ^oC  + 273 =  258 K

The volume is V  =  1000 \  in^3  = \frac{1000}{1728} =  0.5787 ft^3 =  0.0164\ m^3

The initial absolute pressure is P  =  30\  lb/in^2 = 30\  lb/in^2  *  \frac{1}{ \frac{in^2}{144ft^2} } =  4320 \ lb/ft^2 = 308 .19 KPa

Generally from ideal gas equation we have

P *  V  = m RT

Here m is the weight nose wheel tire in pounds

R is the gas constant of air with value R =  0.287 \  \frac{KJ}{kg\cdot K}

So

308 .19  * 0.0164  = m * 0.287* 258

=> m  =  0.068 \  kg

Converting to pounds

m  =  0.068 * 2.205

m  = 0.14994 \  lb

Form this equation P *  V  = m RT specific volume is

\frac{V}{m} =  \frac{RT}{ P}

=>    V_s = \frac{V}{m} =  \frac{0.287 * 258 }{308 .19 }

=> V_s = 0.2403 \  m^3 /kg

4 0
4 years ago
The power in an electric circuit varies inversely with the resistance. If the power is 2,200 watts when the resistance is 25 ohm
valentinak56 [21]

Answer:

The value of resistance when power is 1100 watts = R_{2} = 50 ohms

Explanation:

Power P_{1} = 2200 Watts

Resistance R_{1} = 25 ohms

Power P_{2} = 1100 Watts

Resistance R_{2} = we have to calculate

Given that the power in an electric circuit varies inversely with the resistance

⇒ P ∝ \frac{1}{R}

⇒ \frac{P_{2} }{P_{1} } = \frac{R_{1} }{R_{2} }

⇒ \frac{1100}{2200} = \frac{25}{R_{2} }

⇒ R_{2} = 50 ohms

This is the value of resistance when power is 1100 watts.

6 0
4 years ago
A glucose solution being administered with an IV has a flow rate of 1.86 cm3/min. What will the new flow rate be if the glucose
Sedbober [7]

Answer:

Q' = 0.688 cm³/min

Explanation:

Given that initial flow rate

Q= 1.86 cm³/min

Lets final flow rate is Q' cm³/min

From Poiseuilles law for

Q=\dfrac{\Delta P\pi r^4}{8\eta L}       -------1

Where

ΔP=Pressure difference

L= Length

η=viscosity

r= Radius

Given that all other factor is same only viscosity become 2.7 η.

New viscosity, η'= 2.7  η

New discharge Q'

Q'=\dfrac{\Delta P\pi r^4}{8\eta' L}   ----2

Q' η' = Q  η

Form 1 and 2 equation

Q' x  2.7  η = 1.86 x  η

Q' = 0.688 cm³/min

5 0
3 years ago
What happens if you stick a charged balloon to a metal object?
nikitadnepr [17]
It will have static of course and if you touch it it will shock you
3 0
3 years ago
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