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makkiz [27]
3 years ago
14

Which features on mars point to the possibility of liquid water on the planet?

Physics
2 answers:
Stells [14]3 years ago
7 0
     In 2011 NASA researchers identified features on Mars known as Recuring Slope Lineae ( or RSL ). RSL are relatively dark and narrow features.They are thought to be a signs of salty liquid water on Mars. One hypothesis for RSL formation is that they form when underground bodies of salty water leak into the surface.
   Answer: Recuring Slope Lineae. 
DedPeter [7]3 years ago
5 0

gullies and stream-like channels

Explanation: correct on gradpoint

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What is the displacement of a spring if it has a spring constant of 10 N/m, and a force of 2.5 N is applied?
pochemuha
O.25 m is the displacement
5 0
3 years ago
Read 2 more answers
In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?
agasfer [191]

Answer:

The frequency of the photon is 3.069\times10^{14}\ Hz.

Explanation:

Given that,

Energy E=4.07\times10^{-19}\ J

We need to calculate the energy

Using relation of energy

E_{4}-E_{2}=\Delta E

Where, \Delta E =  energy spacing

4h\nu-2h\nu=4.07\times10^{-19}

\nu=\dfrac{4.07\times10^{-19}}{2h}

Put the value of h into the formula

\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}

\nu=3.069\times10^{14}\ Hz

Hence, The frequency of the photon is 3.069\times10^{14}\ Hz.

4 0
3 years ago
A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o
Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

6 0
3 years ago
A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter toge
Grace [21]

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

6 0
3 years ago
Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

6 0
3 years ago
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