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gizmo_the_mogwai [7]
2 years ago
5

A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled

2.0 × 107 meters and is only 10 meters away from its starting point. What is the numerical value of the satellite's average velocity after that one hour? 3.77 × 10–2 meters/second
Physics
1 answer:
sesenic [268]2 years ago
3 0
Velocity =

(distance between start point and end point, regardless of the route traveled) / (time spent traveling).

That distance (called the "displacement"), is 10 meters, and almost exactly 1 hour is almost exactly 3,600 seconds. So the numerical value of the velocity during that time is

(10) / (3,600) = almost exactly 0.00278 m/s

= 2.78 x 10^-3 m/s.
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The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
The posted speed limit on the road heading from your house to school is45 mi/h, which is about 20 m/s. If you live 8 km (8,000 m
zaharov [31]

Answer:

20.

Explanation:

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6 0
2 years ago
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In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t
gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0
2 years ago
A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas
vodka [1.7K]

Answer:

v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

F=qE

F_p=F_e

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

\frac{m_ev_e^2}{2d}=qE

we can equals these expressions for both proton and electron, because the forces qE are the same:

\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

4 0
2 years ago
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a_sh-v [17]
1. Confident in Reasoning: Douglas is truthful of his reasoning skills to yield good judgments.

2. Analytical: <span>Douglas is habitually alert to potential problems and vigilant in anticipating consequences and trying to foresee short-term and long-term outcomes of being primary care taker for his wife. </span>
7 0
2 years ago
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