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4 years ago

6

An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rop

e that passes over a frictionless pulley (Fig. P4.45), Nick pulls on the loose end of the rope with such a force that the spring scale reads 250 N. Nick's true weight is 320 N, and the chair weighs 160 N. Nick’s feet are not touching the ground. (a) Draw a pair of diagrams showing the forces for Nick and the chair considered as separate systems and another diagram for Nick and the chair considered as one system. (b) Show that the acceleration of the system is upward and find its magnitude. (c) Find the force Nick exerts on the chair.

1 answer:

Oduvanchick [21]4 years ago

5 0

UHHH WHAT? I DONT GET THAT AT ALLOW

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As you move toward a warning siren, the pitch _____.

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As you move toward the siren the pitch should get louder

Explanation:

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1 year ago

Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of (20.0m/s)ı^−(4.0m/s)ȷ^(20.0m/s)ı^−(4.0m/s

Art [367]

Answer:

(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^

Explanation:

Given

W = 0.56N = mg

m = 0.56/g = 0.56/9.8 = 0.057kg

t = 3.00ms = 3.00×10-³s

Impulse is a vector quantity so we would treat it as such

We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that

u = (20.0m/s)ı^−(4.0m/s)ȷ^

ux = 20m/s

uy = – 4.0m/s

F = – (380N)ı^+(110N)ȷ^

Fx = –380N

Fy = 110N

J = impulse = force × time = F×t

So Jx = Fx ×t

Jy = Fy×t

Jx = –380×3×10-³ = -1.14Ns

Jy = 110×3×10-³ = 0.330Ns

Impulse also equals the change in momentum of the body. So

J = m(v–u)

J/m = v – u

V= J/m + u

Vx = Jx/m + ux

Vx = –1.14/0.057 + 20

Vx = -20 + 20 = 0m/s

Vx = 0m/s

Vy= Jy/m + uy

Vy= 0.33/0.057 + (-4.0)

Vy= 5.79 + (-4.0) = 1.79m/s

V = (0m/s)ı^−(1.79m/s)ȷ^

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3 years ago

100 g of Ice at -10°C is added into a

Andrei [34K]

Answer:

The mass of the juice responsible for melting the ice is 949.043 grams.

Explanation:

By the First Law of Thermodynamics, we understand that juice releases heat to the ice, which turns into water under the assumption that interactions between the ice-juice system and surroundings are negligible and energy processes are done in steady-state. Since juice is done with water, its specific heat will be taken as of the water. The process is described by the following formula:

$m_{i} \cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})] + m_{w} \cdot c_{w}\cdot (T_{4}-T_{3}) = 0$ (1)

Where:

$m_{i}$ - Mass of ice, in grams.

$m_{w}$ - Mass of the juice, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of ice, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of the ice-juice system, in degrees Celsius.

$T_{4}$ - Initial temperature of the juice, in degrees Celsius.

If we know that $m_{i} = 100\,g$ , $c_{i} = 2.090\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.18\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334\,\frac{J}{g}$ , $T_{1} = -10\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ , $T_{3} = 10\,^{\circ}C$ and $T_{4} = 20\,^{\circ}C$ , then the mass of the juice is:

$m_{w} = \frac{m_{i}\cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})]}{c_{w} \cdot (T_{3}-T_{4})}$

$m_{w} = \frac{(100\,g)\cdot \left[\left(2.090\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) - 334\,\frac{J}{g} +\left(4.18\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) \right]}{\left(4.180\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C)}$

$m_{w} = 949.043\,g$

The mass of the juice responsible for melting the ice is 949.043 grams.

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3 years ago

What causes fusion to stop within all stars?

puteri [66]

Hello!

When hydrogen runs out in a star, it can no longer perform nuclear fusion where hydrogen nuclei combine to create helium. This will stop fusion.

Hope this helps!

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3 years ago