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2 years ago

Help me pls really struggling with dat

Physics

1 answer:

pogonyaev2 years ago

7 0

So I’m gonna answer the first question on the second page:
For P the force is friction and for Q the force is forward thrust resultant force

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On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as $v_2 = 150 m/s$

since inside air is atmospheric , $v_1 = 0 m/s$

a) By using bernoulli equation between outside and inside of flight

$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh$

$\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$

$\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]$

$\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]$

$\Delta P = 14512.5 Pa = 14.512 kPa$

b) force exerted on window

Area of window $= 25\times 45 = 1125 cm^2 = 0.1125 m^2$

We know that force is given as

$F = P\times A$

$F = 14512.5 \times 0.1125$

F = 1632.65 N

5 0

3 years ago

if the magnitudes of the forces vary with time as F1=Ct and F = 2Ct, where C equals to 7.5 N/s and t is time, find the time t0 a

Degger [83]

Answer:

The tension in the string is equal to Ct

And the time t0 when the rension in the string is 27N is 3.6s.

Explanation:

An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.

So T = Ct

When T = 27N then t = T/C = 27/7.5 = 3.6s

4 0

2 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

3 years ago

When the resistance in a circuit remains constant, how are the voltage and current related?

solmaris [256]

V=IR, therefore when resistance is constant the voltage and current are directly proportional

6 0

3 years ago

Read 2 more answers

Monochromatic light from a distant source is incident on a slit 0.705 mm wide. On a screen 2.13 m away, the distance from the ce

vampirchik [111]

Answer:

492.183 nm

Explanation:

x = Distance from the central maximum to the first minimum = 1.35 mm

l = Distance of screen = 2.13 m

d = Distance of gap = 0.705 mm

m = Order = 1

We have the relation

$tan\theta=\dfrac{x}{l}\\\Rightarrow \theta=tan^{-1}\dfrac{1.35\times 10^{-3}}{2.13}\\\Rightarrow \theta=0.04^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{dsin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.705\times 10^{-3}\times sin0.04}{1}\\\Rightarrow \lambda=4.92183\times 10^{-7}=492.183\ nm$

The wavelength of the light is 492.185 nm

5 0

3 years ago

Read 2 more answers