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3 years ago

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An elevator car, with a mass of 450 kg is suspended by a single cable. At time = 0s, the elevator car is raised upward. The tens

ion on the cable is constant at 5000N during the first 3 seconds of operation. Determine the magnitude of the velocity of the elevator at time = 3 seconds.

1 answer:

Alexeev081 [22]3 years ago

5 0

Answer:

33.33 m/s

Explanation:

m = 450 kg. T = 5000 N, t = 3 seconds,

let the net acceleration is a.

T = m a

a = 5000 / 450 = 11.11 m/s^2

u = 0 , v = ?

Let v be the velocity after 3 seconds.

Use first equation of motion

v = u + a t

v = 0 + 11.11 x 3 = 33.33 m/s

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Answer:

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3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

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Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

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Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

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3 years ago

Which of the following graph is used for determining the instantaneous velocity from the slope?

hoa [83]

Answer:

B. x - t graph

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Generally, the slope of the line of a position-time (x-t) graph is typically used to determine or calculate the velocity of an object.

An instantaneous velocity can be defined as the rate of change in position of an object in motion for a short-specified interval of time. Thus, an instantaneous velocity is a quantity that can be found by measuring the slope of a line that is tangent to a point on the graph.

Hence, the x - t graph also referred to as the position-time graph is used for determining the instantaneous velocity from the slope.

<u>For example;</u>

Given that the equation of motion is S(t) = 4t² + 2t + 10. Find the instantaneous velocity at t = 5 seconds.

Solution.

$S(t) = 4t^{2} + 2t + 10$

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3 years ago

Read 2 more answers

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

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Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

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3 years ago

Read 2 more answers

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Margarita [4]

To solve this problem we will apply the momentum conservation theorem, that is, the initial momentum of the bodies must be the same final momentum of the bodies. The value that will be obtained will be a vector value of the final speed of which the magnitude will be found later. Our values are given as,

$m_1 = 1300kg$

$m_2 = 2400kg$

$u_1 = 12m/s i$

$u_2 = 18m/s j$

Using conservation of momentum,

$m_1u_1+m_2u_2 = (m_1+m_2)v_f$

$1300*12i-2400*18j = (1300+2400)v_f$

Solving for $v_f$

$v_f = 4.2162i-11.6756j$

Using the properties of vectors to find the magnitude we have,

$|v| = \sqrt{(4.2162^2)+(-11.6756)^2}$

$|v| = 12.4135m/s$

Therefore the magnitude of the velocity of the wreckage of the two cars immediately after the collision is 12.4135m/s

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3 years ago