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2 years ago

An object with mass of 4kg is thrown with initial velocity of 20m/s from point A and follows the track of ABCD.

Physics

1 answer:

soldier1979 [14.2K]2 years ago

5 0

<u>The distance of the length YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach the ground is 0.17s</u>

Data given;

Mass = 4kg

Initial velocity = 20m/s

length CD = 5m (from the image given)

<h3>Determine The Length of YD</h3>

$YD=CDsin$ θ = $5sin45=\frac{5}{\sqrt{2} } = 3.57m$

<h3>The velocity of the object at point D</h3>

The change in kinetic energy is given as

Δ in kinetic energy = Δ in potential energy + work done by friction

K.E - 1/2m $v^2$ = mgh $_1$ - mgh $_2$ + (-μmg.x)

K.E = mg(50 - 3.57) + (-mg(0.3*100) + 1/2 m $v^2$

$\frac{1}{2}mv^2_f=mg(46.43)-mg(30)+\frac{1}{2} m(400)\\\\v^2_f=20(16.43)+400\\v^2_f=728.6\\v=\sqrt{728.6} \\v_f=26.99 = 27m/s$

The velocity of the object at D with a distance of 5m.

<h3>The the required for the object to reach ground</h3>

The velocity of the object in the y-axis is

$v_y=vsin45=19.09$

Acceleration in y-axis = 9.8

Height = 3.57m

h = $ut+\frac{1}{2}at^2$

$3.57=19.28(t)+\frac{1}{2}(9.8)t^2\\3.57=19.28t+4.9t^2\\4.9t^2+19.28t-3.57=0\\a=4.9, b=19.28, c= -3.57\\t=\frac{-19.28+\sqrt{(19.28)^2-4(4.9)(-3.57)} }{(2*4.9)} \\t=\frac{-19.28+21.02}{9.8}$

Taking the positive value

$\frac{-19.28+21.02}{9.8}=0.17s$

The time required for the object to reach ground is 0.17s

From the calculations above,<u> the distance YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach ground is 0.17s</u>

Learn more on projectile motion here;

brainly.com/question/1130127

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Density of copper is 8.96 g/cm3and density of zinc is 7.14 g/cm3. Brass is an alloy made of copper and zinc. You are given a sam

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Answer:

The material cost for making one ton of the brass sample that I have is $8149.04.

Explanation:

Density of copper = 8.96 g/cm^3 = 8.96×10^-3 kg/cm^3

Price of copper = $6.13/kg

Price of copper per volume = 8.96×10^-3 kg/cm^3 × $6.13/kg = $0.0549/cm^3

Density of zinc = 7.14 g/cm^3 = 7.14×10^-3 kg/cm^3

Price of zinc = $1.8/kg

Price of zinc per volume = 7.14×10^-3 kg/cm^3 × $1.8/kg = $0.0129/cm^3

Price of brass per volume = 0.0549 + 0.0129 = $0.0678/cm^3

Density of brass I have is 8.32 g/cm^3 = 8.32 g/cm^3 × 1 kg/1000 g × 1 ton/1000 kg = 8.32×10^-6 ton/cm^3

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3 years ago

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

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Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .