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3 years ago

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Nuclear binding energy is necessary to overcome which of the following? Einstein's mass defect

1 answer:

wel3 years ago

6 0

Nuclear binding energy is necessary to overcome EINSTEIN'S MASS DEFECT. Nuclear binding energy refers to the energy required to separate an atomic nucleus into its constituent elements, that is protons and electrons. The mass of a nucleus is always less than the sum of all its constituents. The difference is a measure of the nuclear binding energy which holds the nucleus together and it is called mass defect and can be calculated for using Einstein's formula for mass defect.

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Match the events related to the formation of the universe with the stages during which they occurred.

igomit [66]

Answer:

does this even make sense

Explanation:

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3 years ago

Which statement best describes the atoms of the gas neon?

Leya [2.2K]

Answer:

They:

-Are far from each others

-Move constantly

-Move freely (all directions)

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3 years ago

A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward

riadik2000 [5.3K]

Answer:

Torque, $\tau=34.6\ N.m$

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

$\tau=Fr\ sin\theta$

$\tau=80\times 0.5\ sin(60)$

$\tau=34.6\ N.m$

So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.

7 0

3 years ago

If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the

GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = -10 m

$y_{0}$ : is the initial height = 0

$v_{0_{y}}$ : is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

$t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s$

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

$x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}$

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

$x_{f} = 5.51 m/s*1.43 s = 7.88 m$

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0

2 years ago

In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

6 0

3 years ago