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hichkok12 [17]
3 years ago
15

A 0.76-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one

, and the amount that the spring stretches from its unstretched length triples. What is the mass of the second block?
Physics
1 answer:
sergey [27]3 years ago
4 0

Answer:

1.54 kg

Explanation:

mass of first block (m) = 0.76 kg

acceleration due to gravity (g) = 9.8 m/s

what is the mass (m) of the second block

  • force = kx ⇒ mg = kx

        mg = kx

        where m is the mass, g is the acceleration due to gravity, k is the  

        spring constant and x is the extension

        0.76 x 9.8 = kx

       7.5 = kx

        k = 7.5/x ... equation 1

  • when a second block is attached to the first one the amount of stretch triples (this means that extension (x) = 3x)

        therefore the new mass becomes m + 0.76 and the extension  

        becomes 3x

        with the new mass and extension, mg = kx now becomes

        (m+0.76)g = k(3x) ... equation 2

        Recall that k = 7.5/x from equation 1, substituting this value of k into      

        equation 2 we have

         (m+0.76)g =  \frac{7.5}{x} × (3x)

         (m+0.76)g =  7.5 × 3

          substituting the value of g = 9.8 m/s^{2}

         (m + 0.76) x 9.8 = 7.5 x 3

          m + 0.76 = 22.5 ÷ 9.8

          m + 0.76 = 2.3

          m = 2.3 - 0.76 = 1.54 kg

       

         

       

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Answer:

Yeah  I think you're right

Explanation:

Because every obj is in motion till acted upon by a force(the branch)

4 0
3 years ago
The index of refraction of a clear plastic is listed as 1.89 in the book, but you measured the angle of incidence 63.5° and the
lakkis [162]

Answer:

<h2>index of refraction = 1.69</h2><h2>percentage error = 10.58%</h2>

Explanation:

According to Snell's law, the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. The constant is known as the refractive index.

Mathematically \frac{sin i}{sin r} = n

i = angle of incidence measured = 63.5°

r = angle of refraction measured = 32°

n = refractive index

n = \frac{sin 63.5^{0} }{sin 32^{0} } \\n = \frac{0.8949}{0.5299}\\ n = 1.69

The index fraction calculated is approx. 1.69.

If the index of refraction of a clear plastic as listed in the book is 1.89 and the calculated is 1.69, the percentage error will be calculated as thus;

%error = \frac{1.89-1.69}{1.89} * 100

%error = \frac{0.2}{1.89}*100

%error  = \frac{20}{1.89}

%error  = 10.58%

6 0
3 years ago
9.80 1. The density of mercury is 13600 kg/m³. What is this value in g/cm³? 2. Find the mass of water which will fit in a large
My name is Ann [436]

Answer:

1. 13..6 grams per centimeters cubed.

2. Mass = 400kg

3. Volume = 200cm = 2m

Explanation:

1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.

2. density=\frac{mass}{volume}

1000=\frac{mass}{2 * 1 * 0.2}

1000*0.4=mass

400kg = mass

3. density=\frac{mass}{volume}

0.6=\frac{120}{volume}

volume=\frac{120}{0.6}

volume= 200cm

4 0
2 years ago
Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo
prisoha [69]

Answer:

C) 64lb

Explanation:

use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

Vince Weigh in eart  ⇒  X

our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

140pounds    ⇒   56lb

160pounds    ⇒ X

X=\frac{(160)(56)}{140} =64lb

Vince weigh on the other planet is C) 64lb

5 0
3 years ago
Read 2 more answers
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

Emf = -N\frac{\Delta \phi}{\Delta t}

Where N is the number of turns

\Delta \phi is the change in magnetic  flux

\Delta t is the time interval

The change in magnetic flux is given by

\Delta \phi = \phi _{f} - \phi _{i}

Where \phi _{f} is the final magnetic flux

and \phi _{i} is the initial magnetic flux

Magnetic flux is given by the formula

\phi = BAcos(\theta)

Where B is the magnetic field

A is the area

and \theta is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, \theta = 0^{o}

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

A = \pi r^{2}

∴ A = \pi (0.15)^{2}

A = 0.0225\pi

∴\phi_{i}  = (1.5)(0.0225\pi)cos(0^{o} )

\phi_{i}  = (1.5)(0.0225\pi)

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

The field pointed upwards, that is \theta = 90^{o}. Since cos (90^{o}) = 0

Then

\phi_{f} = 0

Hence,

\Delta \phi = 0- 0.03375\pi

\Delta \phi = - 0.03375\pi

From the question

\Delta t = 2.8 ms = 2.8 \times 10^{-3} s

Here, N = 1

Hence,

Emf = -N\frac{\Delta \phi}{\Delta t} becomes

Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }

Emf = 37.9 V

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0
3 years ago
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