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1 year ago

The shoemaker-near spacecraft explored two asteroids, mathilde and eros. a big difference between them is that:________.

1 answer:

4 0

A big difference between Mathilde and Eros is that Mathilde represents a group of separate rocks whereas Eros is a solid rock.

<h3>What is an asteroid?</h3>

An asteroid is a celestial body composed of different types of rocky materials (generally clay but also silicate) and also metals (especially Fe).

An asteroid may have different sizes and some of them can be huge as dwarf planets that travel across the universe.

In conclusion, a big difference between Mathilde and Eros is that Mathilde represents a group of rocks whereas Eros is a solid rock.

Learn more about asteroids here:

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A car travels 30 km north in 25 min. and 40 km east in 35 min. What is the total time in hours? Be careful to carry over the pro

inna [77]

25+35=60mins so 1.0 hours

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3 years ago

Read 2 more answers

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

8 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0

3 years ago

Pretend you (80 kg) are making repairs on the outside of the International Space Station. You are floating 20 meters away from t

djyliett [7]

Answer:

32 seconds

Explanation:

m1 = 80 kg

m2 = 10 kg

v2 = 5m/s

According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:

$m_{1} v_{1} =m_{2} v_{2} \\80v_{1} =10*5 \\v_{1} = 0.625\ m/s$

Since the space station is 20 meters away, the time taken to reach it is given by:

$t = \frac{20}{0.625}\\t=32\ s$

It takes you 32 seconds to reach the station.

7 0

2 years ago