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Ludmilka [50]
3 years ago
5

YOU GUYS HELPPPPPPPPPPPPPPPP!!!!!! PLEAZEEEE

Physics
2 answers:
Degger [83]3 years ago
5 0
It is Spiral Galaxies
Vinil7 [7]3 years ago
5 0
Gruerjueuudneieijebebe herbeuse je due due isbwi si wushu we www was wso you know ware I just
You might be interested in
For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul
loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

\frac{P_{o} }{P_{1} } =36.73 (M₁ = 3)

If Po₁ = Po₂

\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825

Table A-1:

\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0
3 years ago
A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the
vitfil [10]
Beef and cheddar is your answer
3 0
3 years ago
A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/
Artyom0805 [142]

Answer:

I=182.97\ kg-m^2

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, \tau=860\ N

Angular acceleration of the shell, \alpha =4.7\ m/s^2

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

\tau=I\alpha

I is the rotational inertia of the shell

I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2

So, the rotational inertia of the shell is 182.97\ kg-m^2.

7 0
3 years ago
which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given
Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>
  • Weight of Punch on Earth = 236 lb
  • Desired weight = 118 lb

The mass of Punch will be constant in every planet;

W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}

where;

  • M is the mass of Earth = 5.972 x 10²⁴ kg
  • R is the Radius of Earth = 6,371 km

For Planet Tehar;

g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g

For planet Loput:

g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g

For planet Cremury:

g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g

For Planet Suven:

g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g

For Planet Pentune;

g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g

For Planet Rams;

g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g

The weight Punch on Each Planet at a constant mass is calculated as follows;

W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0
2 years ago
A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
nirvana33 [79]

Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

6 0
3 years ago
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