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Temka [501]
3 years ago
13

calculate the acceleration of Airplane that starts at rest and reaches the speed of 45M/S in nine seconds

Physics
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Acceleration = (Speed or velocity) / Time

= 45/9

= 5 m/s²

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If a plane is inclined at an angle 5o with the horizontal and a cylindrical metal is allowed to roll between two fixed height 92
marysya [2.9K]

Answer:

0.61 second.

Explanation:

So, from the question we are given the following useful parameters or data or information that is going to assist us in solving this question and they are;

(1). Inclination of plane, θ = 50°.

(2). ''The horizontal and a cylindrical metal is allowed to roll between two fixed height 92cm apart on the plane" that is height = 0.92 m(92cm).

(3). The "acceleration of the cylinder to be 2/3 × g Sin (theta) and g = Acceleration due to gravity = 9.8m/s^2

Soz let us delve right into the solution to the question above. We will be making use of the mathematical representation or Equation below;

s = (1/2) a × t².

Where s = height, a = acceleration due to gravity and t = time.

0.92 = (9.8 × sin 50°/ 3) × t².

t² = 0.37

t = √ 0.37.

t =0.61 seconds.

8 0
4 years ago
A nonconducting sphere contains positive charge distributed uniformly throughout its volume. Which statements about the potentia
just olya [345]
This statement about the potential due to the sphere is correct: THE POTENTIAL IS HIGHEST AT THE CENTER OF THE SPHERE.
The potential is highest at the center of the sphere because, charge usually flow from the region of higher potential to lower potential.
7 0
4 years ago
Discuss Joule-Thompson effect with relevant examples and formulae.
Delicious77 [7]

Answer:

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

Explanation:

Joule -Thompson effect

 Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process  

Let

 P_1,T_1 Pressure and temperature at inlet and

 P_2,T_2 Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

\mu _j=\left(\frac{\partial T}{\partial p}\right)_h

Now from T-ds equation

dh=Tds=vdp

So

Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp

⇒dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

So Joule -Thompson coefficient

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

\dfrac{\partial v}{\partial T}=\dfrac{v}{T}

So by putting the values in

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

\mu _j=0 For ideal gas.

6 0
3 years ago
Plz solve this. plz plz plz plz simple machine ​
nalin [4]

Answer:

Explanation:

i.  CW moment = 10 N (10 cm) + 30 N (30 cm) - 60 N (40 cm) = - 1400 N-cm

ii.  ACW momenet = 60 N (40 cm) - 10 N (10 cm) + 30 N (30 cm) = 1400 N-cm

iii.  No. The lever is not balanced in the situation. Because the moment is ± 1400 N-cm.  if balance, the moment must be Zero.

iv.   the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm

take moment from Δ (support)

60(40) = 10(x) + 30(30)

2400 = 10x + 900

10x = 2400 - 900

10x = 1500

x = 1500/10

x = 150 cm  

therefore, the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm

5 0
3 years ago
THIS IS SOOO IMPORTANT PLS HELP I’LL GIVE BRAINIEST
Ket [755]

Answer:

Chemical etching is a process of printed circuit board (PCB) manufacturing that provides  many advantages over mechanical methods.

Explanation:

There's one! UvU hope this helped in whatever you're in for! <3

Good Luck!

6 0
3 years ago
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