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3 years ago

A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the total distance walked by the hiker?

Physics

2 answers:

Nitella [24]3 years ago

5 0

22 km it’s not about how far he traveled from his original starting spot just how far he walked so..22km

AnnyKZ [126]3 years ago

3 0

Answer:

22 km

Explanation:

11km + 11km= 22km

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eimsori [14]

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3 years ago

A amusement park moves riders in a circle at a rate of 6.0m/s if the radius is 9.0 meters what is the acceleration of the ride

oksano4ka [1.4K]

Centripetal acceleration is (speed-squared) / (radius)

CA = (6 m/s)² / (9 m)

CA = (36 m²/s²) / (9 m)

CA = (36/9) (m²/m·s²)

<em>Centripetal acceleration = 4 m/s²</em>

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3 years ago

compare and contrast the medical understanding of death (that you have read about and learned about in this chapter) and the pop

mixer [17]

The medical understanding of death is related to the scientific approach, and the popular understanding is related to the inclusive spiritual and cultural approaches.

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8 0

2 years ago

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o

Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

$v_y = 0$

While the horizontal velocity is

$v_x = 15 m/s$

The horizontal distance covered is

$d_x = 2.7 m$

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

$t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s$

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

$d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m$

8 0

3 years ago

In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y

brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

$E=mc^{2}$

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

$Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules$

Thus the mass that is covertred at 100% efficiency is

$mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg$

Part 2)

At 30% efficiency the mass converted equals

$mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg$

8 0

3 years ago