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Lena [83]
3 years ago
10

A 2.5-kg rock is dropped off a 32-m cliff and hits a spring, compressing it 57 cm. What is the spring constant? Round your answe

r to two significant figures.
Physics
2 answers:
Valentin [98]3 years ago
6 0

The answer is 4800 on the assignment. good luck

Alla [95]3 years ago
6 0

Answer:

The spring constant of the spring is k=4.9\times 10^3\ N/m.

Explanation:

Mass of the rock is 2.5 kg

It is dropped off a 32 m cliff and hits a spring, compressing it 57 cm. It is required to find the spring constant.

When it falls form a height, it possess potential energy and when it hits a spring, the potential energy gets converted to elastic potential energy such that :

mgh=\dfrac{1}{2}kx^2

Here, k is spring constant

k=\dfrac{2mgh}{x^2}\\\\k=\frac{2\cdot2.5\cdot10\cdot32}{0.57^{2}}\\\\k=4924.59\ N/m

k=4.9\times 10^3\ N/m

So, the spring constant of the spring is k=4.9\times 10^3\ N/m.

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slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a
Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:

l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section <em>A</em> at that point:

A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}

And the raising speed <em>v </em>of the water is given by:

v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}

where <em>q</em> is the water flow (1 cubic foot per minute).

7 0
3 years ago
A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s
DedPeter [7]

Answer:

212.8 m/s^{2}

Explanation:

Time taken by stone to cover horizontal distance

t=\sqrt{\frac {2h}{g}} where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

t=\sqrt{\frac {2*2.1}{9.81}}= 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

a=\frac {v^{2}}{r} where r is radius of circle, substituting r with 1.1m

a=\frac {15.3^{2}}{1.1}

a=212.8 m/s^{2}

Therefore, centripetal acceleration is 212.8 m/s^{2}

8 0
3 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete
jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

3 0
3 years ago
Electric eels generate electric pulses along their skin that can stun an enemy. In a test, an electric eel generated pulse of 50
nydimaria [60]

Answer:

The power of the pulse is 40 Watts

3 0
3 years ago
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