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Juli2301 [7.4K]
3 years ago
15

As an object is raised to a higher position,Which type of energy increases?

Physics
2 answers:
sertanlavr [38]3 years ago
8 0
As the object is raised higher, its gravitational potential energy increases which is denoted in a formulae as gpe = mgh where m is mass, g is gravitational accelaration and h is height.
Therefore height increases proportionally to gpe if mass and acceleration are kept constant!

If you need further help over this topic, I will be more than happy to help you out, just message me!

Hope this helps!
Marizza181 [45]3 years ago
7 0
By raising a position of an object, Its potential energy increases.
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When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting
Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0
3 years ago
Which of the following statements is true?
xenn [34]

Correct answer choice is :


C) The freezing and melting temperatures of a substance are the same.


Explanation:


Fluids have a particular temperature at which they convert into solids, identified as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small variations among these measures can be seen. The freezing point of a matter is the same as that substance's melting point. At this distinct temperature, the substance can exist as either a solid or a liquid. At temperatures below the freezing/ melting point, the substance is a solid.


4 0
3 years ago
Read 2 more answers
Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi
77julia77 [94]

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

3 0
3 years ago
One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = 9.11\times 10^{-31}\ kg

B = Magnetic field = 0.044 T

q = Charge of electron = 1.6\times 10^{-19}\ C

The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s

Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

5 0
3 years ago
A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 H
Ganezh [65]

Answer:

<em> 508Hz</em>

Explanation:

A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - this phenomenon is beat production

frequency is the number of oscillation a wave makes in one seconds.

f1-f2=beats

therefore f1=512Hz

f2=?

beats=4Hz

512Hz-f2=4Hz

f2=512-4

f2=508Hz

the original frequency of the violin is 508Hz

6 0
3 years ago
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