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torisob [31]
3 years ago
12

A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

aced with a 2.5 kg mass.
Physics
1 answer:
dolphi86 [110]3 years ago
5 0
In the first case, the force acting on the spring is the weight of the mass:
F=mg=(2.0 kg)(9.81 m/s^2)=19.6N
This force causes a stretching of x=6.4 cm=0.064 m on the spring, so we can use these data to find the spring constant:
k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m

In the second case, the first mass is replaced with a second mass, whose weight is
F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N
And since we know the spring constant, we can calculate the new elongation of the spring:
x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm
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anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change
Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

8 0
3 years ago
What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these
max2010maxim [7]

Answer:

E. all of these

Explanation:

The designation of a point in space all the points that necessary

- reference point

- a direction

- fundamental units

- a direction

- motion

all are necessary to designate a point in space. Hence option E is correct.

For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.  

8 0
3 years ago
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mylen [45]

Answer:

density = 5520 kg/m^3

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given that

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we know,

g = \dfrac{GM}{r^2}

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M = \dfrac{gr^2}{G}

M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}

M = 5.972 x 10²⁴ kg

density =\dfrac{mass}{volume}

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378  x 10³)³

V = 1.08 x 10²¹ m³

density = \dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}

density = 5.52 x 10³  kg/m^3

density = 5520 kg/m^3

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<h3>What is Andromeda Galaxy?</h3>

The Andromeda Galaxy is the nearest spiral neighbor that has spectral lines showing a blue shift.

Therefore, this concludes that, the Universe has stopped expanding. This galaxy is slowly shifting towards us.

Learn more about Andromeda Galaxy.

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