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3 years ago

Where are the nucleus and the electrons located in this atom

Physics

1 answer:

levacccp [35]3 years ago

5 0

In the middle the nucleus is thr brain of it and the electrons spin around the whole atom

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When a ball is launched from the ground at a 45° angle to the horizontal, it falls back to the ground 50 m from the launch point

Inessa [10]

Answer:

Explanation:

Given

angle through which ball is launched $=45^{\circ}$

Range of ball=50 m

Range of projectile is $=\frac{u^2sin2\theta }{g}$

$50=\frac{u^2sin90}{9.8}$

u=22.136 m/s

If ball is thrown straight upward

$v^2-u^2=2as$

$0-(22.136)^2=2(-9.8)s$

$s=\frac{22.136^2}{2\times 9.8}$

s=25 m

(b)For Projectile time of flight is

$t=\frac{2usin\theta }{g}$

$t=\frac{2\times 22.136\times sin45}{9.8}$

t=3.19 s

7 0

3 years ago

A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr

erastova [34]

Gpe = mgh
gpe = 50*10*3
gpe = 1500 J

7 0

3 years ago

How do line symmetry differ from rotational symmetry?.

sergeinik [125]

Answer:

A line of symmetry is a line that separates a shape into two identical halves.
Rotational symmetry is the same thing except when you rotate the object, it has to have the exact same line of symmetry.

<u><em>Hope this helps!!!</em></u>

3 0

3 years ago

Which of the following would add more resistance? Check all that apply.

Blababa [14]

Answer:C. E.

Explanation:

4 0

3 years ago

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

3 years ago