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likoan [24]
3 years ago
12

Suppose that the velocity (in meters per second) of a sky diver falling near the Earth's surface is given by the following expon

ential function, where time is the time after diving measured in seconds. How many seconds after diving will the sky diver's velocity be meters per second? Round your answer to the nearest tenth, and do not round any intermediate computations.

Physics
1 answer:
Mice21 [21]3 years ago
6 0

Answer:

4.7 s

Explanation:

The complete question is presented in the attached image to this solution.

v(t) = 61 - 61e⁻⁰•²⁶ᵗ

At what time will v(t) = 43 m/s?

We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.

43 = 61 - 61e⁻⁰•²⁶ᵗ

- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18

e⁻⁰•²⁶ᵗ = (18/61) = 0.2951

In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205

-0.26t = -1.2205

t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.

Hope this Helps!!!

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A cruise ship is having troubles with buoyancy. What is a reasonable solution? A. Increase the weight of the ship above water B.
Setler [38]

If a cruise ship is having troubles with buoyancy, then spread the weight of the ship over a greater volume.

Answer: Option D

<u>Explanation: </u>

Buoyancy is the upward thrusting phenomenon of water acting on any object immersed partially or fully in water body. Hence, it creates the buoyant forces that is inversely proportionate to the immersing body's density. If the immersing body's density is higher than the density of the immersing medium then the body will get completely immersed in the water.

Similarly, in case of less, the buoyant forces act on the body will prevent it from complete immersion and allow it to float on water. Mostly cruise ships and other navy vessels use this phenomenon to keep on floating on surface of water.

In the present condition, the solution for buoyancy problem faced by a cruise ship can be solved by decreasing the density of the ship. And the ship's density can be decreased by increasing the ship's volume or by spreading the ship's weight over a greater volume.

5 0
3 years ago
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3. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second i
Darya [45]

Answer:

(a) Angular velocity will be 125.6 rad/sec

(b) Linear velocity will be 144.44 m /sec

(c) Centripetal acceleration = 1849.3031 g

Explanation:

We have given diameter d = 2.30 m

So radius r = \frac{d}{2}=\frac{2.30}{2}=1.15m

(a) Speed is given as 1200 rev/min

We know that angular velocity is given by \omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 1200}{60}=125.6rad/sec

(b) Linear speed is given by v=\omega r=125.6\times 1.15=144.44m/sec

(c) Centripetal acceleration is given by a_c=\frac{v^2}{r}=\frac{144.44^2}{1.15}=18141.664m/sec^2

We know that g=9.81m/sec^2

So 18141.66m/sec^2=\frac{18141.664}{9.81}=1849.3031g

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Explanation: The magnitude of the force required to move the electron is given as

F = ma

The force exerted on the charge by the electric field of intensity (E) is given by

F = Eq

Thus

Eq = ma

a = E * q/ m

Where a = acceleration of charge

E = strength of electric field = 7400N/c

q = magnitude of electronic charge = 1.609 * 10^-6c

m = mass of an electronic charge = 9.109 * 10^-31kg

a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31

a = 11906.6 * 10^-16 / 9.019 * 10^-31

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