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2 years ago

Which geologic feature would most likely be represented by contour lines drawn far apart from one another?

Physics

1 answer:

Anna [14]2 years ago

5 0

It would most definitely not be a hill or a cliff. I believe it would be a plain.

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Roller coasters accelerates from initial speed of 6.0 M/S2 final speed of 70 M/S over four seconds. What is the acceleration?

choli [55]

Answer:

$a=16\ m/s^2$

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The formula to calculate the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The roller coaster moves from vo=6 m/s to vf=70 m/s in t=4 seconds. To calculate the acceleration, solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{70-6}{4}=\frac{64}{4}$

$\boxed{a=16\ m/s^2}$

3 0

3 years ago

An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH

Doss [256]

Answer:

Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF

Capacitive reactance = Xc=?

Xc= $\frac{1}{2\pi fC}$

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$

Xl= 2π * f * L

Xl= 2π * 399 * 147 * $10^{-3}$

Xl= 368.5 Ω

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$ = $\sqrt{222^2 + (368.5 - 17.267)^2}$

Z= 415.5 Ω

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

3 0

3 years ago

What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg

scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

$F_c = F_g$

$\frac{mv^2}{r} = \frac{GmM}{r^2}$

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

$v^2 = \frac{GM}{r}$

At the same time we know that the period is equivalent in terms of the linear velocity to,

$T = \frac{2\pi}{\frac{v}{r}}$

$v = \frac{2\pi r}{T}$

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

$v^2 = \frac{GM}{r}$

$( \frac{2\pi r}{T})^2 = \frac{GM}{r}$

$T = \sqrt{\frac{4\pi^2 r^3}{GM}}$

Replacing we have,

$T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}$

$T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})$

$T= 1.74hour$

Therefore the correct answer is C.

7 0

3 years ago

15 points for 2 questions

Elden [556K]

I believe the answer would be c because i think that you multiply the 2

8 0

2 years ago

Read 2 more answers

What is the component in a car engine that would detect the need for air?

Ipatiy [6.2K]

Answer:

Control of air–fuel ratio

Oxygen sensors tell the ECU whether the engine is running rich (too much fuel or too little oxygen) or running lean (too much oxygen or too little fuel) as compared to ideal conditions (known as stoichiometric).

Explanation:

4 0

3 years ago