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Degger [83]
3 years ago
9

Our Milky Way galaxy is 100,000 ly in diameter. A spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1 ly

. What is the speed of the spaceship relative to the galaxy? and How long is the crossing time as measured in the galaxy's reference frame?
Physics
1 answer:
krek1111 [17]3 years ago
3 0

Answer:

t= 100000.000005 years

Explanation:

Our Milky Way galaxy is 100,000 ly in diameter.

the galaxy's diameter to be a mere 1 ly.

We know that

L=L_0\sqrt{1-\frac{v^2}{c^2} }

L= 1 ly

L_0= 100,000 ly

1=100,000\sqrt{1-\frac{v^2}{c^2} }

v= 0.99999999995 c

so time t= \frac{d}{v} = \frac{100000c}{0.99999999995c}

t= 100000.000005 years

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Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

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Explanation:

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In a bicycle dynamo,does 1. The permanent magnet surrounds a conducting coil 2. The conducting coil rotates when the rear wheel
atroni [7]

Explanation:

3: the electricity is generated in the permanent magnet

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What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?
lorasvet [3.4K]

The momentum of the x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and h is Planck's constant,(h=6.62607004 × 10-34<span> m2 kg / s).The momentum is: 2.2 x 10^(-25).</span>

The momentum can be calculated also as: p=mv, where m is the mass of the electron and v is the speed.

So v=p/m,p is known,and also the mass of the electron (m=9.10938356 × 10-31<span> kilograms).</span>

v=2.2 x 10^(-25)/9.10938356 × 10-31<span> kilograms=0.24 x 10^6 m/s</span>

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3 years ago
A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f
Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, F=70\ N

Mass of the dog is, m=25\ kg

Angle of inclination is, \theta =30°

Acceleration due to gravity is, g=9.8\ m/s^2

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

F_g=mg=25\times 9.8=245\ N

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N

Therefore, the normal force acting on the dog is 210 N.

6 0
4 years ago
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