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8090 [49]
3 years ago
12

A 56 kg object is attached to a rope, which can be used to move the load vertically.

Physics
1 answer:
Sergio039 [100]3 years ago
8 0

Answer:

a) T=549.36N Upwards

b) T=448.56N Upwards

c) T=650.16N Upwards

Explanation:

The very first thing we can do to solve this problem is to draw a free body diagram we can use to analyze the situation (see attached picture).

On the diagram we can see there are only two forces acting on the object: the tension of the rope and the weight of the object itself.

a)

Since the object is moving at a constant speed, this means that its acceleration will be zero. So we can do a sum of forces like this:

\sum F=0

T-W=0

T=W

T=mg

T=(56kg)(9.81m/s^{2})

T=549.36N upwards

b)

For part b, since the object is accelerating downwards, we wil say that it's acceleration is negative, so a=-1.8m/s^{2}

so we can do a sum of forces again:

\sum F=ma

so

T-W=ma

T=ma +W

T=ma+mg

T=m(a+g)

and now we substitute:

T=(56kg)(-1.8 m/s^{2}+9.81 m/s^{2})

which yields:

T=448.56N upwards (in this particular case, the tension always goes upwards)

c)

Since the object is moving upwards, we can say that its acceleration will be positive, so a =1.8m/s^{2}

we can take the solved equation we got on the previous part of the problem, so we get:

T=m(a+g)

T =(56kg)(1.8 m/s^{2}+9.81 m/s^{2})

which yields:

T=650.16N upwards

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Answer:

21,000 N

Explanation:

You would use the formula F=ma

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A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a
IgorLugansk [536]

Answer:1.5

Explanation:

Given

mass of first  cart m_1=6 kg

initial Velocity u_1=3 m/s

mass of second cart m_2=3 kg

u_2=0 m/s

In the absence of External Force we can conserve momentum

m_1u_1+m_2u_2=(m_1+m_2)v

v=\frac{m_1u_1+m_2u_2}{m_1+m_2}

v=\frac{6\times 3+3\times 0}{6+3}

v=2 m/s

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K.E._2=\frac{1}{2}(m_1+m_2)v^2

K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2

K.E._2=18 J

Initial Kinetic Energy

K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2

K.E._1=\frac{1}{2}6\times 3^2+0

K.E._1=27 J

ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5

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4 years ago
1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out i
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Acceleration = (change in velocity ( final speed - starting speed))/ (time)

Acceleration = (18-30)/10.5

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Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2

Distance = 252.2 meters

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3 years ago
Temperature is a measure of the average ____________ energy of an object's particles. light mechanical potential kinetic
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Temperature is a measure of the average kinetic energy of the particles of a substance.
Hope this helps! :)
6 0
4 years ago
Read 2 more answers
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s lat
victus00 [196]

Answer / Explanation

It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.

So for proper understanding, the two parts of the question will be highlighted.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?

b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.  

Solve for Vmax

Step Process

a)  Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]

Where V₀ = 6m/s,

We have,

           h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]

                 = 0.411 m

Where V₀ = 9.5m/s

We have,

     h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]

                 = 1152 m

b)  From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.

Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.

Therefore Vₙ = V₀max = 9.8 m/s

4 0
3 years ago
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