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lidiya [134]
4 years ago
14

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

, the object moves across the surface until it encounters a rough incline. The object moves UP the incline and stops a height of 1.5 m above the horizontal surface.
(a) How much work must be done to compress the spring initially?
(b) Compute the speed of the mass at the base of the incline.
(c) How much work was done by friction on the incline?
Physics
1 answer:
Svetradugi [14.3K]4 years ago
8 0

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

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The volume of water is 2500m³ and the fractionof the water is 73.5% of the water contained in that portion of the aquifer would have been removed.

What is aquifer?
An aquifer is a subterranean layer of porous, water-bearing rock, rock fissures, or unconsolidated materials. A water well can be used to obtain groundwater from aquifers. The features of aquifers vary widely. Aquitard is a low-permeability bed along an aquifer, while aquiclude (or aquifuge) is a solid, impermeable region below or covering an aquifer, the pressure of which could form a restricted aquifer.

Area of the unconfined aquifer=10000m².
water table drops by 1m
porosity = 34%
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Volume of water removed= Sy× volume
                                          = 0.25×10000×1
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1 year ago
The main difference between speed and velocity involves
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4 years ago
A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 130 m above the
QveST [7]

Answer:

t =\sqrt{\frac{2*(-100m)}{-9.8 m/s^2}}= 4.518 s

And now we can find the final distance on the x axis using the formula:

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The velocity on x not changes and is the same plane spped V_x = 150 m/s, if we replace we got:

X= 150 m/s * 4.518 s= 677.63 m

Explanation:

For this case we have a illustration for the problem in the figure attached.

We need to find how far short of the target should be the plane, we have the following info given:

v_{ix}= 150 m/s

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First we can find the total time in order to reach the groung using the following kinematic formula:

y_f = y_i + v_{iy} +\frac{1}{2}gt^2

And replacing we have:

-100 m = 0 +0 -\frac{1}{2} (9.8 m/s^2) t^2

And solving for t we got:

t =\sqrt{\frac{2*(-100m)}{-9.8 m/s^2}}= 4.518 s

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D = V_x t

The velocity on x not changes and is the same plane spped V_x = 150 m/s, if we replace we got:

X= 150 m/s * 4.518 s= 677.63 m

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Margarita [4]

Answer:

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