Ask question

Ask question

All categories

3 years ago

12

NASA sends an unmanned lander to test conditions on the surface of Mars. What is the magnitude of the gravitational force acting

on the lander on the surface of Mars? DATA: Mass of Mars = 6.42·1023 kg Radius of Mars = 3397 km Mass of lander = 138.1 kg

1 answer:

kozerog [31]3 years ago

4 0

Answer:

The magnitude of the gravitational force acting on the lander on the surface of Mars is 512.46 N.

Explanation:

The Universal law of gravity is define as:

$F = G \frac{M.m}{R^{2}}$ (1)

Where F is the gravitational force, G is gravitational constant, M is the mass of Mars, m is the mass of the lander and R is the radius of Mars.

Before replacing the values in equation 1 it is necessary to express the radius of mars in terms of meters:

R = $3397 kmx\frac{1000 m}{1 km}$ ⇒ R = $3397000 m$

Finally, equation 1 can be used:

$F = (6.67x10^{-11} N.m^{2}/Kg^{2}) \frac{(6.42x10^{23} kg)(138.1 kg)}{(3397000 m)^{2}}$

$F = 512.46 N$

Hence, the magnitude of the gravitational force acting on the lander on the surface of Mars is 512.46 N.

You might be interested in

An airplane flies with a constant speed of 800km/h. How far can it travel in 2 hours?

Nataly [62]

Because the airplane flies at 800 km/h, it will fly 800 km in 1 hour. In order to find how far it travels in 2 hours, we multiply this number by 2. 800*2 = 1600, so the plane will travel 1600 km in 2 hours.

5 0

3 years ago

What would you have loved to press the pause button on so you could go deeper

kiruha [24]

Answer:

~Banana Fish~

3 0

3 years ago

A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft

erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

$SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)$

Now, differentiating both sides with respect to time 't', we get:

$2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}$

Now, we are given speed as 25 mph. So, $\frac{de}{dt}=25\ mph$

Also, when $e=91\ mi$ , we can find 'x' using equation (1). This gives,

$x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi$

Now, plug in the values of 'e' and 'x' and solve for $\frac{dx}{dt}$ . This gives,

$\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph$

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

3 years ago

A 26-kg sled is on a snow-covered slope. The coeﬃcients of friction between the sled’s runners and the snow are µs = 0.096 and µ

sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction $\mu _s=0.096$

coefficient of kinetic friction $\mu _k=0.072$

In order to move sled from rest we need to provide a force greater than static friction which is given by

$f_s=\mu mg=0.096\times 26\times 9.8=24.46 N$

After Moving Sled kinetic friction comes in to play which is less than static friction

$f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N$

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

3 0

3 years ago