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3 years ago

4. How do ordinary objects behave in microgravity?

Physics

1 answer:

Nikolay [14]3 years ago

8 0

Answer:

Bubbles paused

Explanation:

the air bubble doesn't rise because it is no lighter than the water around it—there's no buoyancy. The droplet doesn't fall from the leaf because there's no force to pull it off. It's stuck there by molecular adhesion.

for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth

On Earth, the buoyancy of the air bubbles causes them to rise to the top together, creating a segregation between air and water. However, in microgravity, nothing forces the air bubbles to interact and thus rise together, Green said.

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2. Find the velocity when the displacement is 40 m and the time is 5 s.

LenKa [72]

Answer:

V=?

S= 40m

t=5s

V=S/t

V=40/5

V=8m/s

3 0

3 years ago

A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend

GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

$a=\frac{h}{5(t_1)^2}$

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = $y = \frac{1}{2}a(t_1)^2$

Velocity = $v_1=at_1$

<h3>Stage 2</h3>

Constant velocity where

Velocity = $v_o=v_1=at_1$

Distance =

<h3> $y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\$ </h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = $v_0=v_1=at_1$

Distance =

$y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2$

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = $\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2$

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

$a=\frac{h}{5(t_1)^2}$

6 0

3 years ago

Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons

gregori [183]

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

8 0

3 years ago

Oceanic crust is much denser than continental crust

Free_Kalibri [48]

Oceanic because it’s denser

8 0

2 years ago

Read 2 more answers

A golf club with 65J of kinetic energy strikes a stationary golf ball with a mass of 46g. The energy transfer is only 20% effici

umka21 [38]

Kinetic energy of golf club = 65J,
kinetic energy supplied to golf ball = 20% of 65 = 0.2 * 65 = 13J,
kinetic energy of ball = [mass * Velocity²]/2,
mass = 46gm = 0.046Kg,
[0.046 * V²]/2 = 13, or 0.046 *V² = 26,
V² = 26/0.046 = 565.22,
V = 23.77 m/sec = initial velocity of golf ball after hitting.

4 0

3 years ago