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3 years ago

An astronaut lands on a new planet. The new planet’s mass is half that of the

1 answer:

7 0

Ah hah ! Very nice problem.

-- Since the mass is 1/2 of the Earth's mass, the gravitational forces between him
and the new planet would be 1/2 of what they are on Earth.

-- Since the distance between him and the center of the new planet is 1/2 of the
distance between him and the Earth's center, the gravitational forces between him
and the new planet would be (2)² = 4 times what they are on Earth.

-- These conditions combine to make his weight (1/2) x (4) = 2 times his earth weight.

-- He weighs 1,000 N (about 224.8 pounds) on the new planet.

(Maybe that's why they selected an astronaut for this mission who only weighs
about 112.5 pounds on Earth.)

You might be interested in

1 point

Tju [1.3M]

Answer:

Option A nuclear

Explanation:

The rate of electricity production in nuclear power plant is much higher as compared to the rate of electricity generation in gas, wind and solar power plants.

Thus, in case where large amount of electricity is to be produced in a short period then one must rely on nuclear power plants.

Therefore, option A is correct

7 0

3 years ago

It is your parent's 25th Wedding Anniversary and you thought it would be special to throw them a surprise party. Your entire fam

KonstantinChe [14]

Answer:

120 miles per hour.

Explanation:

We need to find the time it takes my parents to drive home from the cottage. Since my father drives at 60 miles per hour, and the cottage is 240 miles from our home, and distance = speed × time. So, time = distance/speed = 240 mi/60 mi/h = 4 h.

So, it will take my father 4 hours to drive home from the cottage.

Since I have 2 hours to prepare for the party, the time left for me to drive to the cottage is 4 - 2 hrs = 2 hrs.

So, I'm supposed to drive to the cottage in at most 2 hours.

The speed at which I must drive in this time period is thus, speed = distance/time = 240 miles/2 hours = 120 miles per hour.

So, I must drive at a minimum speed of 120 miles per hour.

4 0

3 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0

3 years ago

What is an ellipse?

Rom4ik [11]

Answer:

i think it's C thx correct me if wrong

6 0

2 years ago

A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0

3 years ago