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elena-s [515]
3 years ago
5

You run completely around a 400m track in 80s. What was your average velocity?

Physics
1 answer:
dalvyx [7]3 years ago
6 0

Answer:

V=?

S=400m

t=80s

V=S/t

V=400/80

V=5m/sec

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The answer is A. A folkway is a closely held belief by a specific group of people.
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An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach
matrenka [14]

Answer:

H = 34.43 m

Explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection, \theta=60^{\circ}

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m

So, the maximum height reached by the object is 34.43 m.

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3 years ago
An electron is moving horizontally east in an electric field that points vertically downward. The electric force on the electron
saveliy_v [14]

Answer:

d. )directed upward.

Explanation:

As the electron has a negative charge, when under the influence of an electric field, is subject to an electric force, which direction is the opposite to the direction of the electric field.

This is because the electric field has the same direction that the force on a positive test charge at the same point.

As the electric field points vertically downward, the electric force on the electron (a negative charge) points vertically upward.

So, the statement d. is the one that results to be true.

7 0
3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

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3 years ago
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What is the current in mA through a 500 Ω resistor that is connected to a 1.5 V battery? Show all calculations using Ohm’s law (
Ludmilka [50]

Answer:

3 mA.

Explanation:

The following data were obtained from the question:

Resistor (R) = 500 Ω

Potential difference (V) = 1.5 V

Current (I) =.?

Using the ohm's law equation, we can obtain the current as follow:

V = IR

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Divide both side by 500

I = 1.5 / 500

I = 3×10¯³ A.

Therefore, the current in the circuit is 3×10¯³ A.

Finally, we shall convert 3×10¯³ A to milliampere (mA).

This can be obtained as follow:

Recall:

1 A = 1000 mA

Therefore,

3×10¯³ A = 3×10¯³ × 1000 = 3 mA

Therefore, 3×10¯³ A is equivalent to 3 mA.

Thus, the current in mA flowing through the circuit is 3 mA.

3 0
3 years ago
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