2 years ago

Which of the following are features of the nucleus of an atom

Physics

1 answer:

vichka [17]2 years ago

4 0

We r made of atom so v can’t touch anything hehe I just joking

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Someone help please i’ll give brainliest

Fantom [35]

Directly proportional to volume, according to charles’s law

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3 years ago

Technological advances are now making it possible to link visible-light telescopes so that they can achieve the same angular res

leonid [27]

Answer: The angle of resolution =47.05°

Explanation:

The angle of resolution represents the resolve power and precision of optical instruments such as the eye, camera and e en a microscope. It is given by the equation:

Sin A = 1.220(W÷D)

Sin A= 1.220(300÷500)

Sin A = 1.220(0.6)

Sin A = 0.732

A= Sin^-1 0.732

A= 47.05°

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2 years ago

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<img src="https://tex.z-dn.net/?f=%5Chuge%5Cbold%7B%5Cpurple%7B%5Cbold%7B%E2%9A%A1Gravitational%20Constant%3F%E2%9A%A1%7D%7D%7D%

baherus [9]

$\huge\underline\mathtt\colorbox{cyan}{G=}$

$6.673 \times {10}^{ - 11}$

And unit is Nm^2/kg^2

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2 years ago

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A 600-g block is dropped onto a relaxed vertical spring that has a spring constant k =190.0 N/m. The block becomes attached to t

charle [14.2K]

Answer:

Work done will be 2.205 j

Explanation:

We have given that the spring is compressed b 37.5 cm

So d = 0.375 m

Mass of the block m = 600 gram = 0.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Gravitational force on the block $F=mg=0.6\times 9.8=5.88N$

Now we know that work done is give by $W=Fd=5.88\times 0.375=2.205J$

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2 years ago

A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa

kotykmax [81]

Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is $v_i=180~m/s$ , the final velocity is $v_i=0~m/s$ , and the total time of the motion is $\Delta t=0.02~s$ , so the acceleration is given by
$a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2$
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
$S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m$
So, the bullet penetrates the sandbag 1.8 meters.

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2 years ago