Modern space suits augment the basic pressure garment with a complex system of equipment and environmental systems designed to keep the wearer comfortable, and to minimize the effort required to bend the limbs, resisting a soft pressure garment's natural tendency to stiffen against the vacuum. A self-contained oxygen supply and environmental control system is frequently employed to allow complete freedom of movement, independent of the spacecraft.
Three types of spacesuits exist for different purposes: IVA (intravehicular activity), EVA (extravehicular activity), and IEVA (intra/extravehicular activity). IVA suits are meant to be worn inside a pressurized spacecraft, and are therefore lighter and more comfortable. IEVA suits are meant for use inside and outside the spacecraft, such as the Gemini G4C suit. They include more protection from the harsh conditions of space, such as protection from micrometeorites and extreme temperature change. EVA suits, such as the EMU, are used outside spacecraft, for either planetary exploration or spacewalks. They must protect the wearer against all conditions of space, as well as provide mobility and functionality.
Answer:
a) T = 608.22 N
b) T = 608.22 N
c) T = 682.62 N
d) T = 533.82 N
Explanation:
Given that the mass of gymnast is m = 62.0 kg
Acceleration due to gravity is g = 9.81 m/s²
Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.
So;
To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;
T = mg
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs the rope at a constant rate tension in the string is
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs up the rope with an upward acceleration of magnitude
a = 1.2 m/s²
the tension in the string is T - mg = ma (Since acceleration a is upwards)
T = ma + mg
= m (a + g )
= (62.0 kg)(9.81 m/s² + 1.2 m/s²)
= (62.0 kg) (11.01 m/s²)
= 682.62 N
When the gymnast climbs up the rope with an downward acceleration of magnitude
a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)
T = mg - ma
= m (g - a )
= (62.0 kg)(9.81 m/s² - 1.2 m/s²)
= (62.0 kg)(8.61 m/s²)
= 533.82 N
Answer:
– 2.5 m/s²
Explanation:
We have,
• Initial velocity, u = 180 km/h = 50 m/s
• Final velocity, v = 0 m/s (it stops)
• Time taken, t = 20 seconds
We have to find acceleration, a.
a = (v ― u)/t
a = (0 – 50)/20 m/s²
a = –50/20 m/s²
a = – 5/2 m/s²
a = – 2.5 m/s² (Velocity is decreasing) [Answer]
Answer:
hi mate,
interesting question, first of all the pressure is determined by using the following formula:
Pg = p * G * h
where p is the density of the liquid, G is the gravity and h is the height difference, in you case you have:
p = 1015 kg/m3
G = 9.8m/s2
h = 0.085 m
insert these values into the equation above:
Pg = 1015 kg/m3 * 9.8m/s2 * 0.085 m = 849.81 kg·m-1·s-2 or 849.81 pascal
hope it helps, :-)
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