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Flura [38]
3 years ago
7

The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for

his business. Every day, his route takes him 22.5 miles against the current and back to his dock, and he needs to make this trip in a total of 9 hours. He has a boat in mind, but he can only test it on a lake where there is no current. How fast must the boat go on the lake in order for it to serve the ferry operator’s needs?
Physics
1 answer:
user100 [1]3 years ago
7 0
The first step is to determine the rate using the given values. Distance formula 9(x-6)=45 is used where the rate is x-6. Value of x is equivalent to 11. The average speed that covers 45 miles in 9 hours is 5 miles per hour having no current.
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A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra
blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

y=\sqrt{1+x^3}

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s

7 0
3 years ago
Direct current, DC, flows in a _________ direction while alternating current, AC, the direction of flow ___________. A) single,
kramer
The flow of Direct current (DC) is constant and flows in one direction. Most digital electronics make use of DC. Alternating current (AC) periodically flows in reverse and is mostly used to deliver power to houses, buildings and the like. With that alone, you can already rule out A, C and D. 

The answer would then be B. constant, periodically reversing. 
4 0
3 years ago
Read 2 more answers
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

5 0
3 years ago
When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

\theta = 54.6^{\circ}

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s

Now,

The angular velocity can be calculated as:

v = \omega R

Thus

\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s

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3 years ago
If a runner has a speed of 8.66m/s and runs for 46.2s what distance is covered? tv = d
kati45 [8]

Answer:

\text{Using the formula: }v=\frac{d}{t}\\\therefore vt=d\\\text{Plug and chug:}46.2(8.66)=400.092\text{ metres}

6 0
3 years ago
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