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2 years ago

A man walks 6 km north, then turns and walks 6 km east what is the resultant and angle

1 answer:

5 0

That would depend on the man’s starting latitude. If 6.7 mi south of the north pole, for example, the resultant would be 6.7 mi due north of the starting point.

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Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

3 years ago

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

3 years ago

The amount of fluid displaced by a submerged object depends on its

Daniel [21]

The amount of fluid displaced by a submerged object depends on its volume.

7 0

3 years ago

Read 2 more answers

If the mass of an object is 8 kg and its momentum is -80 kgm/s, what is its velocity?

Dimas [21]

An object's momentum is the product of its mass and its velocity:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

p = -80kg×m/s

m = 8kg

Plug in these values and solve for v:

-80 = 8v

v = -10m/s

Choice D

4 0

3 years ago

As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?

liubo4ka [24]

As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.

The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.

The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.

A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.

Learn more about capacitor here:-brainly.com/question/14883923

#SPJ4

8 0

1 year ago