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podryga [215]
3 years ago
7

Define mass defect and binding energy​

Physics
1 answer:
natima [27]3 years ago
6 0

Answer:

About Transcript. Nuclear binding energy is the energy required to split an atom's nucleus into protons and neutrons. Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. The binding energy of a system can appear as extra mass, which accounts for this difference.

Explanation:

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A 10 cm diameter pulley is used to lift a bucket of cement weighing 400 N. How much force must be applied to the rope to lift th
Alina [70]

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────┈⊰

\large \bold {ANSWER}

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

\large \bold {EXPLANATION}

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, <u>because generally when it is mobile, this information is highlighted in the question</u>.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

<u>Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.</u>

3 0
2 years ago
In both the camera and the __________, light enters a narrow opening and is projected onto a photosensitive surface. Group of an
Citrus2011 [14]

Answer: The HUMAN EYE

Explanation:

The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.

In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.

The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.

The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.

5 0
3 years ago
A Ground Fault Circuit Interrupter (GFCI) works by comparing the amount of current going to and returning from equipment along t
DanielleElmas [232]
Choice-B is the correct one.
5 0
3 years ago
Read 2 more answers
A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the
ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

a_{B} =a_{D}  +(a_{B/D} )_{t}

2.5j=1.5j  +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

2\alpha =1\\\alpha =0.5rad/s^{2}CW

b) The acceleration of point A is:

a_{A} =a_{D}  +(a_{A/D} )_{t}

(aA/D)t = ADαj

a_{A} =a_{D}  +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}

The acceleration of point E is:

(aE/D)t = -EDαj

a_{E} =a_{D}  -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}

7 0
4 years ago
A charge of 19 nC is uniformly distributed along a straight rod of length 15 m that is bent into a circular arc with a radius of
Salsk061 [2.6K]

Answer:

The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C

Explanation:

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8 0
3 years ago
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