All categories

3 years ago

A 400-kg space probe has a weight of 3,560 N on one of the above planets. According to the table

Physics

1 answer:

Rzqust [24]3 years ago

8 0

Answer:

Venus

Explanation:

To know the correct answer to the question, we shall determine the weight of the space probe on each planet to see which will correspond to 3560 N.

This is illustrated below:

For Mercury:

Mass = 400 Kg

Gravitational force = 3.7 N/Kg

Weight =?

Weight = mass × gravitational force

Weight = 400 × 3.7

Weight = 1480 N

For Venus:

Mass = 400 Kg

Gravitational force = 8.9 N/Kg

Weight =?

Weight = mass × gravitational force

Weight = 400 × 8.9

Weight = 3560 N

For Earth:

Mass = 400 Kg

Gravitational force = 9.8 N/Kg

Weight =?

Weight = mass × gravitational force

Weight = 400 × 9.8

Weight = 3920 N

For Neptune:

Mass = 400 Kg

Gravitational force = 11 N/Kg

Weight =?

Weight = mass × gravitational force

Weight = 400 × 11

Weight = 4400 N

For Jupiter:

Mass = 400 Kg

Gravitational force = 23.1 N/Kg

Weight =?

Weight = mass × gravitational force

Weight = 400 × 23.1

Weight = 9240 N

SUMMARY:

Planet >>>>>> Weight

Mercury >>>> 1480 N

Venus >>>>>> 3560 N

Earth >>>>>>> 3920 N

Neptune >>>> 4400 N

Jupiter >>>>>> 9240 N

From the above calculations, we can see that the weight of the space probe (i.e 3560 N) correspond to the weight on venus (i.e 3560 N)

You might be interested in

A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di

Sidana [21]

Answer:

(a) $\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) $\theta = 85.44^{\circ}$

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

$\hat{A} = \frac{\vec{A}}{|A|}$

$\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}$

$\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}$

$\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) To calculate the angle between the two vectors say A and A' is given by:

$\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta$

$\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})$ (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

$\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$

Now,

Using equation (1) :

$\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})$

$|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261$

Thus

$\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})$

$\theta = cos^{- 1}(0.07946) = 85.44^{\circ}$

4 0

3 years ago

A distant galaxy has a redshift z = 5.82 and a recessional velocity vr = 287,000 km/s (about 96% of the speed of light.) Notice

vlabodo [156]

Answer: 4100 Mpc

Explanation:

Since H o = 70 km/s/Mpc

Redshift z = 5.82

Recessional velocity vr = 287,000 km/s

Then, the distance to the galaxy in light years will be:

= Recessional velocity / H o

= 287000 / 70

= 4100 Mpc

7 0

3 years ago

A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels

spin [16.1K]

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

7 0

3 years ago

What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k

Molodets [167]

Answer:

$\dfrac{I_1}{I_2}=12.25$

Explanation:

$r_1$ = 14 km

$r_2$ = 49 km

Intensity of a wave is inversely proportional to distance

$I\propto \dfrac{1}{r^2}$

So,

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25$

The ratio of the intensities is $\dfrac{I_1}{I_2}=12.25$

6 0

3 years ago

When is a model used in science?

Mamont248 [21]

C. Both of the above

8 0

3 years ago