All categories

3 years ago

How would a person's mass be affected by traveling to Saturn and why?

1 answer:

7 0

A person's mass would not change before, during, or after a trip to Saturn, unless he gained or lost some weight as a result of unrelated eating habits, because mass doesn't depend on an object's location, gravitational environment, color, cost,
texture, temperature, political affiliation or gender orientation.

You might be interested in

A group of hikers hear an echo 3.3 s after they shout. The temperature is 20◦C.

romanna [79]

Answer:

560 m

Explanation:

The speed of sound in air is approximately:

v ≈ v₀ + 0.6T

where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.

The speed of sound at 20°C at that altitude is:

v ≈ 327 + 0.6(20)

v ≈ 339 m/s

The sound travels from the hikers to the mountain and back again, so it travels twice the distance.

339 m/s = 2d / 3.3 s

2d = 1118.7 m

d = 559.35 m

Rounding, the mountain is approximately 560 m away.

4 0

3 years ago

How long is a string under 240 N of tension whose mass is 0.086 kg if a wave travels through it at a speed of 12 m/s?

Darina [25.2K]

Answer:

The length of the string is 0.051 meters

Explanation:

It is given that,

Tension in the string, T = 240 N

Mass of the string, m = 0.086 kg

Speed of the wave, v = 12 m/s

The speed of the wave on the string is given by :

$v=\sqrt{\dfrac{T}{M}}$

M is the mass per unit length of the string i.e. M = m/l.......(1)

So, $M=\dfrac{T}{v^2}$

$M=\dfrac{240\ N}{(12\ m/s)^2}$

M = 1.67 kg/m

The length of the string can be calculated using equation (1) :

$l=\dfrac{m}{M}$

$l=\dfrac{0.086\ kg}{1.67\ kg/m}$

l = 0.051 m

So, the length of the string is 0.051 meters. Hence, this is the required solution.

6 0

3 years ago

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

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

if i release one steel ball from the top of a ramp and the other ball from the 40cm mark will they have the same acceleration?

solniwko [45]

Answer:

Yes, they will also have the same acceleration. Acceleration is controlled by the amount of weight (def. amount of gravitational pull on a given object) that the ball has. For one ball to accelerate quicker than the other, it would need a propulsion element, which it does not.

Explanation:

6 0

2 years ago

If 5kg Stone and 1kg stone throw the from the building which will land more fa ster and why?

sweet [91]

Answer:

Both stones will land at the same time because both stones will fall with the same acceleration through the same height.

Explanation:

We are given that

Mass of stone ,m1=5 Kg

Mass of stone, m2=1 kg

We have to find which stone more faster will land and why.

$h=u+\frac{1}{2}gt^2$

Initial velocity of both stones=0

$h=\frac{1}{2}gt^2$

$t^2=\frac{h}{g}$

$t=\sqrt{\frac{h}{g}}$

$t_1=t_2=\sqrt{\frac{h}{g}}$

Because both stones are thrown from the same height.

Both stones will land at the same time because both stones will fall with the same acceleration through the same height and the acceleration does not depend of its mass.

5 0

3 years ago