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Marta_Voda [28]

2 years ago

13

yellow cab taxi charges a $1.75 flat rate in addition to $0.65 per mile. kim has no more than $10 to spend on a ride. how many m

iles can kim travel without exceeding his limit?

1 answer:

Nat2105 [25]2 years ago

5 0

Answer:

you could go 12 miles paying $7.80 and $1.75

So in total being $9.55

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What affects the amount of potential energy stored in the magnetic field when a magnet is moved against a magnetic force?

LekaFEV [45]

Answer: Strength of magnet and distance from magnetic material

Explanation:

The potential energy of a magnet is determined by the strength of the magnet and the distance between a magnet and another magnet or a magnetic material. Magnetic materials are materials that would be attracted when brought close to a magnet, example of magnetic materials are most metals.

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2 years ago

A plane travels 350 km south along a straight path with an average velocity of 125 km/h to the south. The plane has a 30 minute

faust18 [17]

Average Velocity = Total Displacement / Total time

1st part of journey, 350 km at velocity 125 km/h

Time = 350 / 125 = 2.8 hours.

2nd part of journey, 220 km at velocity 115 km/h

Time = 220 / 115 = 1.9 hours

Average Velocity = Total Displacement / Total time

= (350 + 220) / (2.8 + 1.9)

= 570 / 4.7 ≈ 121.3 km/hr

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Option C.

6 0

2 years ago

This is a science Question. Please help.

sattari [20]

Answer:

B

Explanation:

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3 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

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It is called a warm-up.

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3 years ago

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