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2 years ago

8

A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of 0.075 kg. During one weighing, th

e spring is stretched a distance of 12 cm from equilibrium. During a second weighing, the spring is stretched a distance of 18 cm.
9. How much greater is the elastic potential energy of the stretched spring during the second weighing than during the first weighing?

1 answer:

mr Goodwill [35]2 years ago

8 0

elasticity stretches and can also return to it's normal size ..

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I need help!!! 60 POINTS!!!!

Scrat [10]

Correct one is b
Good luck

3 0

3 years ago

Read 2 more answers

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago

The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh

iogann1982 [59]

Answer:103 pounds

Explanation:

Given

width of window $b=35 in.$

height of window $h=20 in.$

standard atmospheric pressure $P_{outside}=14.7 psi$

Also $P_{inside}=1.01P_{outside}$

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

$F_{net}=(P_{inside}-P_{outside})\cdot A$

$F_{net}=P_{outside}(1.01-1)\times 35\times 20$

$F_{net}=14.7\times 0.01\times 35\times 20$

$F_{net}=102.9\ pounds\approx 103\ pounds$

8 0

3 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0

3 years ago

Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in p

irina1246 [14]

Answer:

1 ohm

Explanation:

since there are two identical resistors, one resistor will be

R = $\frac{4}{2}$ =2ohm [ proven as in series $R_{e} = 2 + 2 = 4ohm$ ]

to calculate the equivalent resistance when in parallel:

$\frac{1}{R_{e} } = \frac{1}{R_{1} } + \frac{1}{R_{2}}$

so,

$\frac{1}{R_{e} } = \frac{1}{2} + \frac{1}{2}$

$R_{e} = 1ohm$

4 0

2 years ago