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Mrac [35]
3 years ago
5

Why is it difficult to drop a probe like Galileo? How did engineers solve this problem?

Physics
1 answer:
icang [17]3 years ago
4 0

Explanation:

Galileo spacecraft was sent to orbit around and study Jupiter. It was launched on October 18, 1989 and entered into Jupiter's orbit on December 7, 1995. It had a lot of firsts in its name and most important of them was the first probe that entered in Jovian atmosphere. The spacecraft carried a probe named <em>Galileo</em> which separated from the spacecraft 5 months before it was to reach Jupiter. The separation was done in such a way that the rendezvous of probe and spacecraft with Jupiter coincided.

The mission was very challenging. Some of them are listed below:

1. Because of Jupiter's immense gravity the speed of the probe reached up to 47 KM/Sec as it approached Jupiter, but to enter into Jupiter's atmosphere it had to be brought down to subsonic level (0.34 KM/Sec). Due to this it had to withstand 228G deceleration.

2. Also, the time of probe entry had to be synchronized with that of the rendezvous of spacecraft with Jupiter. Very precise calculations were needed to achieve this.

3. Jupiter has a very thick atmosphere which would result in very high pressure and temperature for anything entering into it atmosphere. A very strong heat shield was needed to withstand such high pressure and temperature. Out of the 339 kg of the probe 152 kg was just the weight of the heat shield. Around 80 kg of the heat shield was lost during the entry. Its instruments were so sturdy that they worked till a pressure of 23 atm and temperature of 153°C.

4. The deceleration of the probe was further aided by using a parachute during the entry.

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It is the force applied to a machine
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after the car leaves the platform , gravity causes it to accelerate downward at a rate of 9.8 m/s2. what is the gravitational fo
alexandr1967 [171]

The gravitational force on the car is the force popularly known
as the car's "weight".  Its magnitude is

            (9.8 m/s²) times (the car's mass, in kilograms) .

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5 0
3 years ago
Riding a bicycle is this kind of long-term memory ______________
Schach [20]

answer

procedural memory ❤

8 0
3 years ago
uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work i
Lesechka [4]

Answer:

(a) The work done is 0.05 J

(b) The  force will stretch the spring by 3.8 cm

Explanation:

Given;

work done in stretching the spring from 30 cm to 45 cm, W = 3 J

extension of the spring, x = 45 cm - 30 cm = 15 cm = 0.15 m

The work done is given by;

W = ¹/₂kx²

where;

k is the force constant of the spring

k = 2W / x²

k = (2 x 3) / (0.15)²

k = 266.67 N/m

(a) the extension of the spring, x = 37 cm - 35 cm = 2 cm = 0.02 m

work done is given by;

W = ¹/₂kx²

W = ¹/₂ (266.67)(0.02)²

W = 0.05 J

(b) force = 10 N

natural length L = 30 cm

F = kx

x = F / k

x = 10 / 266.67

x = 0.0375 m

x = 3.75 cm = 3.8 cm

Thus a force of 10 N will stretch the spring by 3.8 cm

7 0
4 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

8 0
3 years ago
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