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3 years ago

7

You want to use a rope to pull a 10-kg box of books up a plane inclined 30∘ above the horizontal. The coefficient of kinetic fri

ction is 0.29. The rope pulls parallel to the incline.
What force do you need to exert on the rope if you want to pull the box with a constant acceleration of 0.50 m/s2 up the plane?

1 answer:

fgiga [73]3 years ago

5 0

Answer: 78.61N

Explanation

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Each time the heart beats,what does it do to the blood

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Your blood runs through your blood stream, every vein in your body. without that heartbeat the blood would not make it to the required places it needs to run to and you will lose oxygen and parts of your body and organs will shut down.

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3 years ago

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

$P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})$

$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

$V=3.475ft^3/s$

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

$m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s$

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

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2 years ago

The stimuli for kinesthesis is the __________ energy of joint and muscle movement. A. thermal B. electrical C. mechanical D. che

lilavasa [31]

The answer to this question is The first option, Or what I should say "A.Thermal"

Your welcome!

4 0

3 years ago

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Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the

kati45 [8]

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

$U=k\dfrac{q^2}{d}$

In this system with three charges which are equidistant from each other

$U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}$

$\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J$

The potential energy of the system is 0.12959085 J

6 0

3 years ago

A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At th

GREYUIT [131]

Here the block has two work done on it

1. Work done by gravity

2. Work done by friction force

So here it start from height "h" and then again raise to height hA after compressing the spring

So work done by the gravity is given as

$W_g = m_A g(h - h_A)$

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path

$W_f = -(\mu m_A g cos\theta)(\frac{h}{sin\theta} + \frac{h_A}{sin\theta})$

$W_f = -\mu m_A g cot\theta(h + h_A)$

Total work done on it

$W = m_A g(h - h_A) - \mu m_A g cot\theta(h + h_A)$

So answer will be

None of these

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3 years ago

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