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Studentka2010 [4]
3 years ago
12

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How man

y radians did the disk turn while stopping ? C. how many revolutions? ​
Physics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = Δω/t = (0 - π)/3 = π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π radians

θ = 1.5π rad/2π rad/rev = 0.75 rev

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NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
3 years ago
Helppppppppppp im dieing
Otrada [13]
It would be 1. B 2. A 3. A
7 0
3 years ago
What is the magnitude of the gravitational force acting on the earth due to the sun?
expeople1 [14]

Answer: 3.524(10)^{22}N

Explanation:

According to Newton's law of Gravitation, the force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the product of their masses and inversely proportional to the square of the distance:

F=G\frac{(m1)(m2)}{r^2}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}  

m1=1.99(10)^{30}kg is the mass of the Sun

m2=5.972(10)^{24}kg is the mass of the Earth

r=1.50(10)^{11}m  is the distance between the Sun and the Earth

Substituting the values in (1):

F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}   (2)

Finally:

F=3.524(10)^{22}N   This is the gravitational force acting on the earth due to the sun

3 0
3 years ago
Read 2 more answers
The distance between planets varies as they move in their orbits. The minimum distance from the Earth to Mars is about 0.35 AU.
Alina [70]

Answer:

52,360,000km

Explanation:

To solve this problem you use a conversion factor.

By taking into account that 1UA = 1.496*10^{8}km you obtain:

0.35UA*\frac{1.496*10^{8}km}{1UA}=52,360,000km

hence, 0.35UA is about 52,360,000km. This is the least distance between Mars and Earth

5 0
3 years ago
Read 2 more answers
3. A model rocket is launched straight upward at 58.8 m/s.
kolezko [41]

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

   v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t =  \frac{58.8}{9.8}

 = 6 seconds

b). Total time of flight = 2(Time taken to go up)

                                    = 2×6

                                    = 12 sec

c). Formula to get the peak height is,

   h=ut-\frac{1}{2}gt^2

   h = (58.8)6 - \frac{1}{2}(9.8)(6)^2

      = 352.8 - 176.4

      = 176.4 meters

8 0
4 years ago
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