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sergejj [24]
4 years ago
7

Velocity differs from speed in that velocity indicates a particle's __________ of motion. Enter the letter from the list given i

n the problem introduction that best completes the sentence.
Physics
1 answer:
nexus9112 [7]4 years ago
3 0

Before solving this problem first we have to understand the basic of velocity and speed.The speed is a scalar quantity which means the rate of change of distance covered by a particle .As the distance traveled  is a scalar quantity as it has no specific directions,so is the speed.It has simply magnitude.

then we have to understand the velocity.Velocity is a vector quantity.IT is defined as the rate of change displacement of the particle or we may say the speed in a given direction .As we know that displacement is the minimum distance traveled by a particle which is nothing else except the straight line joining two points,so it has direction.hence velocity has a magnitude as well as direction.one has remember that the magnitude velocity is always smaller than or  equal to the speed velocity\leq speed

so the answer will be DIRECTION i.e velocity indicates the direction of motion .

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A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi
bogdanovich [222]

Answer:

1) W_{object} = 400 N

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

\sum F_{Tot} = 0

In this case we just have y-direction forces.

\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0

Now, let's solve the equation for W(object).

W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N

2 ) To find the position of the heaviest weight we need to use the torque definition.

\sum \tau = 0

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

L = 1.50 m  ; total length of the bar.

D_{AP} = x  ; distance between Tension A and P point.

D_{BP} = L-x ; distance between Tension B and P point.

D_{W_{bar}P} = \frac{L}{2}-x ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0

Finally we just need to solve it for x.

x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}

x = 0.28 m

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

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4 years ago
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7 0
3 years ago
A student is about to sled down a hill. He stands on a hill which is 30 meters tall and he weighs 40kg what is his GPE
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Answer:

11760J

Explanation:

Given parameters:

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g is the acceleration due to gravity

h is the height

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