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Alika [10]
4 years ago
10

WILL MARK BRAINLIEST!!!!!!!!!!!!

Physics
1 answer:
PSYCHO15rus [73]4 years ago
4 0
40 miles/hour

Details:
after one hour they have traveled 40mi. So they have traveled 40mi. in 1hr, or 40mi/1hr or 40mi/hr.

the problem is about reading a graph. the graph says that they traveled 40 miles in the first hour

formula for average speed:
miles traveled/time it took=speed or d/t=r
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Water boiling<br><br> Which one is shown? (Look at pic)
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Answer:

conduction.

Explanation:

Hoped I helped! Im Eve btw have a great day and consider marking this brainliest if you do thank you in advanced!

4 0
3 years ago
The _______ theory states that the Moon formed as the result of a collision between Earth and a Marssized object.
FromTheMoon [43]
D.) giant impact

hope this helps 
4 0
3 years ago
Identify the theory that can be used to explain each phenomenon.
ahrayia [7]

Answer:

Diffraction: Wave Theory

Interference: Wave Theory

Reflection: both particles and wave theories

Refraction: both particles and wave theories

3 0
3 years ago
Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0
BaLLatris [955]

Answer:

Part a)

r = 6.96 \times 10^6 m

Part b)

F_g = 3.004 \times 10^3 N

Part c)

a = 0.88 m/s^2

Part d)

v = \frac{2480}{2} = 1240 m/s

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

v = 2480 m/s

now we will have

v = \sqrt{\frac{GM_{mars}}{r}}

now we have

M_{mars} = 6.4191 \times 10^{23} kg

R_{mars} = 3.397 \times 10^6 m

now we have

2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}

r = 6.96 \times 10^6 m

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

F_g = \frac{GM_{mars} m}{r^2}

F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}

F_g = 3.004 \times 10^3 N

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

a = \frac{F_g}{m}

a = \frac{3004}{3400}

a = 0.88 m/s^2

Part d)

As we know by III law of kepler

\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}

here we know that T2 = 8 T1

(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}

\frac{r_1}{r_2} = (\frac{1}{2})^2

so we have

r_2 = 4r_1

as we know that speed is given as

v = \sqrt{\frac{GM}{r}}

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

v = \frac{2480}{2} = 1240 m/s

7 0
4 years ago
Which statement is true about an airplane wing during flight?
allochka39001 [22]
C. air above the wing travels faster than air below the wing.
3 0
4 years ago
Read 2 more answers
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