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jeka94
3 years ago
14

Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l

ight-years away from Earth. Kepler-10b is a superearth, meaning it is a little larger and more massive than Earth. It has a surface gravity that is 1.53 times that of Earth. How much thrust would be required to accelerate a 5,000-kg spacecraft 2.13 m/s2 from the surface of Kepler-10b
Physics
1 answer:
OleMash [197]3 years ago
8 0

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\

<u>F = 85696.5 N = 85.69 KN</u>

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A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc
katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

cos \theta = \frac {V_{river}}{V_{boat}}

When solving for theta, we get that:

\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})

so now we can substitute the corresponding values:

\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})

Which yields:

\theta = 60^{o}

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

V_{y}=5.60km/hr*cos(210^{o})

V_{y}=-4.85km/hr

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

t=\frac{y}{V_{y}}

t=\frac{5.60km}{4.85km/hr}=1.155hr

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

V_{ds}=V_{river}+V{boat}

V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr

and now up stream:

V_{us}=V_{boat}-V{river}

V_{us}=5.60km/hr-2.80km/hr=2.80km/hr

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

t_{ds}=\frac{x}{v_{ds}}

t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr

and the time up stream:

t_{us}=\frac{x}{v_{us}}

t_{us}=\frac{2.80km}{2,80km/hr}=1hr

so the total time will be:

t_{ds}+t_{us}=0.333hr+1hr=1.333hr

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

t=\frac{d}{v_{boat}}

t=\frac{5.60km}{5.60km/hr}

so

t= 1hr

4 0
4 years ago
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What is power?
Blizzard [7]

Answer:

D

Explanation:

P=Work/Time

The rate at which work is done matches that.

6 0
4 years ago
Need help 8th grade science test need help
jolli1 [7]
ANSWER: d, average speed
7 0
2 years ago
Read 2 more answers
As sound waves move from air to glass, the wave property of ________ will cause the wave to bend in that direction
Zarrin [17]
<span>The amplitude because that controls the height of the wave. Correct answer: Amplitude.</span>
5 0
4 years ago
Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, t
makvit [3.9K]

Answer:

A) 30 s, 792 m

B) 10.28 s, 4108.2 m = 4.11 km

Explanation:

A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

Using the equations of motion.

v = u + at

v = final velocity = 120 km/h

u = initial velocity = 70 km/h

a = acceleration = 6000 km/h²

t = ?

120 = 70 + 6000t

6000t = 50

t = (50/6000) = 0.0083333333 hours = 30 seconds.

Using the equations of motion further,

v² = u² + 2ax

where x = horizontal distance covered by the car during this time

120² = 70² + 2×6000×x

12000x = 120² - 70² = 9500

x = (9500/12000) = 0.79167 km = 791.67 m = 792 m

B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

Bullet A is fired upwards with velocity 450 m/s

Bullet B is fired upwards with velocity 600 m/s too

Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.

y = ut + ½at²

For the bullet A

u = initial velocity = 450 m/s

a = acceleration due to gravity = -9.8 m/s²

y = 450t - 4.9t² (eqn 1)

For the bullet B, fired 3 seconds later,

u = initial velocity = 600 m/s

a = acceleration due to gravity = -9.8 m/s²

t = T

y = 600T - 4.9T²

At the point where the two bullets pass each other, the vertical heights covered are equal

y = y

450t - 4.9t² = 600T - 4.9T²

But, note that, since T starts reading, 3 seconds after t started reading,

T = (t - 3) s

450t - 4.9t² = 600T - 4.9T²

450t - 4.9t² = 600(t-3) - 4.9(t-3)²

450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)

450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1

600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0

179.4t - 1844.1 = 0

t = (1844.1/179.4) = 10.28 s

Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.

y = 450t - 4.9t²

= (450×10.28) - (4.9×10.28×10.28)

= 4,108.2 m = 4.11 km

Hope this Helps!!!!

6 0
3 years ago
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