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3 years ago

8

The closest approach distance between Mars and Earth is about 56 million km. Assume you can travel in a spaceship at 58,000 km/h

, which is the speed achieved by the New Horizons space probe that went to Pluto and is the fastest speed so far of any space vehicle launched from Earth. How long would it take to get to Mars at the time of closest approach?

1 answer:

vladimir1956 [14]3 years ago

6 0

Answer:

40.22 days

Explanation:

Given data:

Closest approach distance between Mars and Earth = 56 million km = 56 × 10⁶ km

Speed of the spaceship = 58000 km/h

Now, the time (t) is calculated as:

time = Distance / speed

on substituting the values, we get

t = 56 × 10⁶ km / (58000 km/h)

or

t = 965.517 hours

or

t = 965.517 / 24 days = 40.22 days

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A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

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Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

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w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

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3 years ago

A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of 0.075 kg. During one weighing, th

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elasticity stretches and can also return to it's normal size ..

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2 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

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Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

So

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

2)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

So

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

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2 years ago

Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.

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By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

Tension is also a force having Newton as S.I unit.

The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using Lami's theorem.

The sum of two given angles = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

$\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}$

Where

$\alpha = \beta$ = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

$\frac{T}{sin\alpha } = \frac{W}{sinY}$

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

T = 48.56 N

Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

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2 years ago

In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor

VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

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$I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2$

by perpendicular axis theorem

$I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2$

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2 years ago