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4 years ago

Which force prevents protons from repelling each other inside a nucleus?

Physics

1 answer:

Stella [2.4K]4 years ago

4 0

Answer:

So, the correct answer is the strong nuclear force. It actually pulls together nuetrons and protons that are in the nucleus. At very tiny distances only, like those inside the nucleus, so, this strong force succeded in dealing with the electromagnetic force, and it basically stops the electrical repulsion of protons from blowing apart the nucleus.

Mark as brainlies please, I need a few more :D

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Define habitat preservation and give at least two examples.

lisov135 [29]

Habitat preservation is when people get together or just one person helps to stop the end of a living space for animals such as the monterey aquarium the help sea life by doing research to see why animals are dying and do what they can to stop it. another one if nation parks or forests you are not allowed to harm them in any way they are protected by laws(the government)

7 0

4 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

4 years ago

The rainforest is hot, humid, and has frequent rainfall. In what way do the atmosphere and the hydrosphere contribute to this ty

il63 [147K]

Answer:

The atmosphere which contains nitrogen, oxygen, carbon dioxide, and other gases essential for life, contributes to the rainforest climate. The rainforest is hot, humid and helps balance the amount of carbon dioxide in the atmosphere by absorbing it and releasing more oxygen which is important for climate change.

The hydrosphere, which is the combination of various types of water sources and types found in air, in the form of water vapor; water found below the earth's surface and above, also contributes to the rainforest climate. The frequent rainfall dictates the vegetation growth in that area which also creates weather patterns due to the contribution of the atmosphere and the hydrosphere. The hydrosphere is important in the formation of rain. A rainforest of frequent rainfall indicates a large amount of the combination of various types of water sources and types found in a particular area, which release the water back onto the rainforest.

7 0

3 years ago

A lifeguard on a beach observes that waves have a speed of 2.60 m/s and a distance of 2.50 m between wave crests. What is the pe

Alja [10]

Answer:

T = 0.96 seconds

Explanation:

Given that,

The speed of wave, v = 2.6 m/s

The distance between wave crests, $\lambda=2.5\ m$

We need to find the period of the wave motion. Let T be the period. So,

$v=\dfrac{\lambda}{T}\\\\T=\dfrac{\lambda}{v}\\\\T=\dfrac{2.5}{2.6}\\\\T=0.96\ s$

So, the period of the wave motion is equal to 0.96 seconds.

4 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

F = m a

a = F / m

We use kinematics to find lips

v = v₀ + a t

v = v₀ + (F / m) t

The moment is defined by

p = m v

The moment change

Δp = m v - m v₀

Let's replace the speeds in this equation

Δp = m (v₀ + F / m t) - m v₀

Δp = m v₀ + F t - m v₀

Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

8 0

3 years ago