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3 years ago

Why is the solar system commonly represented as a scale model?

1 answer:

6 0

The actual size of the Solar system is too big to show without making a much smaller model. If someone wants to see the orientation of the planets a model has to be made so we can see it without flying out too space.

You might be interested in

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

8 0

3 years ago

A +6.33 uC charge q1 is attracted by a force of 0.115 N to a second charge q2 that is 1.44 m away. What is the value of q2?

jarptica [38.1K]

Answer:-4.1858

Explanation: I hate Acellus

7 0

2 years ago

An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De

gayaneshka [121]

We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
$F=T-W$
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
$a= \frac{F}{m}$
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
$F=T-W=6000 N-5000 N=1000 N$
To find m, we can use the weight of the system. In fact, the weight of an object is given by
$W=mg$
where $g=9.81 m/s^2$ . Solving for m, and using W=5000 N, we find
$m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg$

and at this point, we can calculate the acceleration of the system (elevator+people):
$a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2$
and the acceleration has the same direction of the resultant force, so upward.

8 0

3 years ago

Which has greater value, a newton of gold on earth or a newton of gold on the moon?

jek_recluse [69]

The moon, because the acceleration due to gravity is less.

4 0

2 years ago

A heavy rope, 80 ft long and weighing 32 lbs, hangs over the edge of a building 100 ft high. how much work w is done in pulling

allochka39001 [22]

The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft

6 0

3 years ago