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3 years ago

A hiker travels 300 m [N], turns around and hikes 550 m[S], and then finds out he has to hike back another 50.0 m

Physics

1 answer:

bulgar [2K]3 years ago

4 0

Answer:

900 m

Explanation:

Given that,

A hiker travels 300 m [N], turns around and hikes 550 m[S], and then finds out he has to hike back another 50.0 m [N].

We need to find the total distance travelled by the hiker.

Let north be positive and south be negative direction.

Distance = total path covered

D = 300 + 550 + 50

D = 900 m

So, the required distance is equal to 900 m.

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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.

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3 years ago

It is August 1st and you are at a Science Camp in Florida. During an outdoor science quiz, you are asked to identify the tempera

KonstantinChe [14]

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3 years ago

A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?

STatiana [176]

Power = Force * Distance/ time
P = 1,250 * 2/3
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So, your final answer is 833.33 Watts

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4 years ago

Which of the following elements do living things have that volcanic rocks do not have?

jek_recluse [69]

Answer: silicon,or maybe none.

Explanation: I searched it up not really sure sorry

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3 years ago

Arocket launches at an angle of 33.6 degrees from the horizontal at a

babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

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Initial velocity of the rocket is, $u=58.5$

A vector at an angle $\theta$ with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle $\theta$ with the horizontal, then the horizontal and vertical components are given as:

$A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})$

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

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Plug in the given values and solve for $u_y$ . This gives,

$u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})$

Therefore, the Y component of initial velocity is 32.37.

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3 years ago