Ask question

Ask question

All categories

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?

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

You might be interested in

A tree frog leaping upward off the tree branch is pulled downward by

jeka94

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