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Y_Kistochka [10]
2 years ago
6

Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle? they are high

enough above the earth's surface that they have escaped the pull of earth's gravity. the space shuttle has anti-gravity devices on it that allow the astronauts to float inside it. the astronauts are falling at the same rate as the space shuttle as it orbits around earth. orbiting objects do not experience any gravity?
Physics
1 answer:
Maslowich2 years ago
7 0

If we see the forces on the astronauts then there is only one force on each astronaut while they are in air.

This force is due to gravity of earth.

Now while the astronauts are in air and doing some activities then the net force on them is counterbalanced on them by centrifugal force.

so we can say the rate of fall of astronauts due to gravity is at same rate as the orbiting rate of the space shuttle.

this is given by

w^2R = g

so all astronauts will experience the situation of free fall.

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John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho
Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

F_{down} L_1 = W_{snow} L_2

so here we have

F_{down} = \frac{L_2}{L_1} (W_{snow})

so here we can say that upward force by which we push up is always more than the downwards force

8 0
2 years ago
How many ternary strings of length 2 n are there in which the zeroes appear only in odd-numbered positions?
Rashid [163]

In a string of length 2n, there are exactly n odd-numbered posiitons. In the remaning n even-numbered positions, we want to choose from only two possible digits, either a 1 or a 2. This means there are 2^n such ternary strings.

8 0
2 years ago
A historical society is testing an old cannon. They
sveta [45]

Answer:

26.2 m/s

Explanation:

We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is

u_y = u sin \theta (1)

where u is the initial speed and \theta = 45.0^{\circ} is the angle of projection.

We can therefore use the following suvat equation for the vertical motion of the ball:

v_y = u_y + at

where v_y is the vertical velocity at time t, and a=g=-9.8 m/s^2 is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:

t=\frac{3.78}{2}=1.89 s

And at the maximum height, the vertical velocity is zero:

v_y=0

Substituting these values, we find the initial vertical velocity:

u_y = v_y - at = 0-(-9.8)(1.89)=18.5 m/s

And using eq.(1) we now find the initial speed:

u=\frac{u_y}{sin \theta}=\frac{18.5}{sin 45.0^{\circ}}=26.2 m/s

4 0
2 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
2 years ago
If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of e
ryzh [129]

Answer:

σ = 1.09 mm

Explanation:

Step 1: Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

Step 2: calculate length of the rod, L

K = \frac{A*E}{L}K=

L

A∗E

L = \frac{A*E}{K}L=

K

A∗E

A=\frac{\pi d^{2}}{4}A=

4

πd

2

d = 20-mm = 0.02 m

A=\frac{\pi (0.02)^{2}}{4}A=

4

π(0.02)

2

A = 0.0003 m²

L = \frac{A*E}{K}L=

K

A∗E

L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=

55X10

6

(0.0003142)∗(200X10

9

)

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

\sigma = \frac{F*L}{A*E}σ=

A∗E

F∗L

\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=

(0.0003142)∗(200X10

9

)

(60X10

3

)∗(1.14)

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

5 0
2 years ago
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