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Levart [38]
3 years ago
9

As a hiker in Glacier National Park, you are looking for a way to keep the bears from getting at your supply of food. You find a

campground that is near an outcropping of ice from one of the glaciers. Part of the ice outcropping forms a slope of angle θup to a vertical cliff. You decide that this is an ideal place to hang your food supply, as the cliff is too tall for a bear to reach it. You put all of your food into a burlap sack, tie an unstretchable rope to the sack, and tie another bag full of rocks to the other end of the rope to act as an anchor. The mass of rocks that you put into the anchor bag is equal to the mass of food in the other bag.What will be the acceleration aof the food bag when you let go of the anchor bag? Assume that the weight of the rope is negligible, and that the ice can be considered frictionless.Express the resulting acceleration aof the food bag in terms of θ and g, the acceleration due to gravity. Let the positive direction for this bag be downwards.
Physics
1 answer:
LenKa [72]3 years ago
8 0

Answer: the acceleration is (g( 1 - sinθ)) / 2

Explanation:

the tension in one bag is expressed as;

T = ma + mg sin θ ........................lets say equ 1

now the tension in another bag is expressed as;

T = -ma + mg..................lets say equ 2

so equate both equations

ma + mg sinθ = -ma + mg

2ma =  mg ( 1 - sinθ)

a = (g( 1 - sinθ)) / 2

therefore the acceleration is (g( 1 - sinθ)) / 2

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6 0
3 years ago
A car starts from the state of xestIf its velocity becomes 70 km/hr in 6 minutes, i) what is the accordine acceleration of F the
Alexxandr [17]

Answer: 3.5\ km

Explanation:

Given

Car starts from the state of rest and acquires a velocity of 70\ km/hr in 6 minutes

Final velocity in m/s is v=70\approx 19.44\ m/s

Using equation of motion

v=u+at\\\Rightarrow 19.44=0+a(6\times 60)\\\Rightarrow a=0.054\ m/s^2

Distance covered in 360 s

\Rightarrow v^2-u^2=2as\\\Rightarrow 19.44^2-0=2\times 0.054\times s\\\Rightarrow s=3500.64\ m\approx 3.5\ km

4 0
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stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

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6 0
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Answer:

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They must be equal to mantain equilibrium on the body and he can stay floating, this force is equivalent to the weight of water displaced

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6 0
3 years ago
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