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Ivahew [28]
3 years ago
15

I am pretty sure its the last one.

Physics
2 answers:
yuradex [85]3 years ago
8 0

Answer:

Has no moons

Explanation:

mars has 2 moons named phobos and deimos

CaHeK987 [17]3 years ago
4 0
The answer is has no moons. Mars has two moons
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When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coeffic
Nuetrik [128]

Answer:

8.1\cdot 10^{-4} C^{-1}

Explanation:

The volumetric expansion of the liquid is given by

\Delta V=\alpha V_0 \Delta T

where

\alpha is the coefficient of volume expansion

V_0 is the initial volume

\Delta T is the change in temperature

For the liquid in this problem,

V_0 = 2.35 m^3\\\Delta T=48.5^{\circ}C\\\Delta V=0.0920 m^3

So we can solve the equation to find \alpha:

\alpha=\frac{\Delta V}{V_0 \Delta T}=\frac{(0.0920 m^3)}{(2.35 m^3)(48.5^{\circ}C)}=8.1\cdot 10^{-4} C^{-1}

7 0
4 years ago
CAN U HELP PLZZ? ;(
Ahat [919]
Answer would be D sorry if wrong
7 0
3 years ago
An object moves in a circle with a period of 0.025 hours. What is its frequency in Hz?
Sedaia [141]

Answer:

40 Hz

Explanation:

f = 1/T = 1 / 0.025 = 40 Hz

7 0
3 years ago
Calculate the density of a sample of gas with a mass of 30 g and volume of 7500 cm3.
zepelin [54]

By definition we have that the density is given by:

D = \frac{M}{V}

Where,

M: mass of the sample

V: volume occupied by the sample

Therefore, substituting values in the given equation we have:

D = \frac{30}{7500}

D = 0.004 \frac{g}{cm^3}

Answer:

the density of a sample of gas with a mass of 30 g and a volume of 7500 cm3 is:

D = 0.004 \frac{g}{cm^3}

4 0
3 years ago
Read 2 more answers
Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.
Ad libitum [116K]

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{4.0 m/s}{6.0 m/s})=arctan(0.67)=33.7^{\circ}

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=\frac{S_y}{v_y}=\frac{360 m}{4.0 m/s}=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

7 0
4 years ago
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