1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vovangra [49]
3 years ago
12

Consider two identical springs. At the start of an experiment, Spring A is already stretched out 3 cm, while Spring B remains at

the zero position. Both springs are then stretched an additional three centimeters. What conclusion can you draw about the force required to stretch these springs during the experiment
Physics
1 answer:
aksik [14]3 years ago
7 0

Answer:

Explanation:

In this interesting exercise we have that spring A is 3 cm longer, due to previous experiments if these experiments did not reach the non-linear elongation point, the cosecant Km of the spring must remain the same, therefore when we lengthen the two springs these the longitudinal are lengthened.

As a consequence of the above according to Hockey law, the prediction of lengthening is the same, therefore the outside is the same in two two systems

            F = K Δx

You might be interested in
An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at th
AlexFokin [52]

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × 10^{-6} m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top  t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = 10^{5} Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

so

pressure at bottom P2 = 10^{5} + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × 10^{5}  Pa

so from gas law

\frac{P1*V1}{t1} = \frac{P2*V2}{t2}

here p is pressure and v is volume and t is temperature

so put here value and find v1

\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

6 0
3 years ago
The density of ice is 0.93 g/cm3. what is the volume, in cm3, of a block of ice whose mass is 5.00 kg? remember to select an ans
serious [3.7K]

The Volume of the ice block is 5376.344 cm^3.

The density of a material is define as the mass per unit volume.

Here, the density of ice given is 0.93 g/cm^3

Mass of the ice block  given is 5 kg or 5000 g

Now calculate the volume of the ice block

density=mass/volume

0.93=5000/Volume

Volume =5376.344 cm^3

Therefore the volume of  ice block is 5376.344 cm^3

7 0
3 years ago
Which of the following must ALWAYS be equal to the buoyant force on an object?
professor190 [17]

Answer:

A

Explanation:

I’m not sure but I hope this helped

8 0
3 years ago
Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg
n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N

Thus, the net force over the body is:

F=(913.14N)\hat{i}+(274.87N)\hat{j}

Next, you calculate the magnitude of the force:

F=\sqrt{(913.14N)+(274.87N)^2}=953.61N

and the direction is:

\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°

7 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
Other questions:
  • While discussing the effects of loose wheel bearings: Technician A says that vibration may occur while driving. Technician B say
    12·1 answer
  • A slinky is stretched across a classroom and moved up and down at a frequency of 2 hz. if the corresponding wave velocity is 4.2
    10·1 answer
  • If a steady magnetic field exerts a force on a moving charge, that force is directed
    13·1 answer
  • Baseballs pitched by a machine have a horizontal velocity of 30 meters/second. The machine accelerates the baseball from 0 meter
    9·1 answer
  • How big is the universe I get it only God knows because he created all of it meaning everything.That is real.
    7·1 answer
  • A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w
    8·1 answer
  • PLEASE HELP ME!!!!!!!!!!
    12·1 answer
  • A wave traveling througil air meets an object and bounced off its flat surface
    15·2 answers
  • An impulse of 12.2kg m/s is delivered to an object whose initial momentum is 4.5kgm/s. What is the object's final momentum?
    12·1 answer
  • WHO IS GOOD AT PHYSICS, ELECTRIC CIRCUITS?​
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!