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Toluwa F.
7+ Years Tutoring, Physics Grad, Biomedical Engineer at 3D4MD :)
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Applied Mathematics
TutorMe
Question:

Using applied mathematics, how long would it take to cook a 20 lb turkey if it takes 3 hours to cook a 15 lb turkey?

Toluwa F.

Using proportionality we know it takes 4 hours, but I need to use applied mathematics! Let it take x hours to cook 20 lb turkey. We know that it takes 3 hours to cook a 15 lb turkey and the time taken to cook is directly related to the weight of the turkey. $$\frac{20}{15} = \frac{x}{3}$$ $$x= 4$$ The 20lb turkey will take 4 hours to cook.

Computer Science (General)
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Question:

How does a router differ from such devices as repeaters, bridges and switches?

Toluwa F.

A Router is a device which is used when multiple devices are needed to connect to the internet using the same Internet Protocol (IP) address. Routers forward data packets from one place to another and forward data depending upon the network address. Very little filtering of data is done through routers. It works on the Network Layer. A Repeater receives a weak signal, amplifies it and then retransmits it. It removes the unwanted noise in an incoming signal. It works in the Physical Layer. A Bridge is a device that connects two LANs. It simply forwards packets without analyzing messages. Bridges are faster than routers but also less versatile. It works in Data Link Layer. The Switch is a device that filters and forwards packets between LAN segments. It works in the Data Link Layer. A switch is a secure device.

Physics
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Question:

The Moon's distance from the Earth varies between $$3.65x10^5 km$$ at perigee and $$4.07x10^5 km$$ at apogee. What is the ratio of it's orbital speed around Earth at perigee to that at apogee?

Toluwa F.

From conservation of angular momentum, we should have $$mv_p r_p = mv_a r_a$$ Then the ratio of orbital speed at perigee to that at apogee is $$\frac{v_p}{v_a} = \frac{r_a}{r_p}$$ Given the radius of orbit at perigee $$r_d = 3.56x10^5m$$ And the Radius of orbit at apogee $$r_a = 4.07x10^5m$$ then $$\frac{v_p}{v_a} = \frac{(4.07x10^5m)}{(3.56x10^5m)}$$ $$= 1.14$$

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