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3 years ago

Select the correct answer from each drop-down menu.

Physics

1 answer:

earnstyle [38]3 years ago

7 0

Answer:

a transverse (sort of a plot of a sine or cosine graph, basically)

b longitudinal

c Electromagnetic (an electric wave and a magnetic wave travelling together at right angles to each other)

Explanation:

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Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

Second part

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

Which of the following is an example of how humans can increase biodiversity?

Leto [7]

Answer:

The following is an example of how humans can increase biodiversity

Provide Wildlife Corridors and Connections Between Green Spaces

Use Organic Maintenance Methods and Cut Back On Lawns

Use a Native Plant Palette and Plant Appropriately

Utilize Existing Green Space Connections

Be Mindful of Non-Native Predators

7 0

3 years ago

A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o

maxonik [38]

For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
v = y*f = 1.5*220 = 330 m/s

4 0

3 years ago

The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?

Greeley [361]

On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where
I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore 
I = 1/12m*2² = 1/3m kg*m² 

The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so 
ω = 2π * 2 rev/s = 4π s^(-1) 

The angular momentum would therefore be 
L = Iω 
= 1/3m * 4π 
= 4/3πm kg*m²/s, where m is the rod's mass in kg. 

The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. 

Edit: 650 g = 0.650 kg, so 
L = 4/3π(0.650) kg*m²/s 
≈ 2.72 kg*m²/s

4 0

3 years ago

One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th

alisha [4.7K]

Answer:

1/2 M V^2 = .1 M g H where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

7 0

2 years ago