All categories

3 years ago

WILL GLADLY GIVE BRAINIEST FOR THIS EASY QUESTION

2 answers:

5 0

A copper ion with a charge of +2 has 29 protons and 27 electrons. Since copper has an atomic number of 29, all atoms of copper will have 29 protons.

Since the number of protons of an element doesn't change. Copper, for example, has two isotopes, copper-63 and copper-65. Copper-63 has 29 protons and a mass number of 63.

Please award me with the BRAINIEST

Alekssandra [29.7K]3 years ago

4 0

The atomic number of Copper is 29. So immediately we know that there are 29 protons in the nucleus of each Copper atom, and 29 electrons in the cloud surrounding the nucleus of each [neutral] Copper atom.

Copper has two stable isotopes ... ⁶³Copper and ⁶⁵Copper. So if the Copper atom you're holding in your hand is not radioactive, then it has either (63-29)=34 neutrons or (65-29)=36 neutrons in its nucleus. (There's never a need to round the number of neutrons, because there's no such thing as a part of a neutron.)

You might be interested in

Please answer these questions, its for a physics assignment

laiz [17]

A no . answer is velocity of the object. B no.answer is Acceleration of the object. C no.answer is straight line shape and a velocity graph with a horizontal shape D no.answer is curved shape and a velocity graph with a straight shape.

3 0

3 years ago

Air is heated in a glass bottle. The heat energy added to the air is 2.0 × 104 joules. What is the change in internal energy of

Liono4ka [1.6K]

The change in internal energy of the gas is $\Delta U = 2.0 \cdot 10^4 J$ .

In fact, the 1st law of thermodynamics states that the change in internal energy of a system is equal to the amount of heat given to the system (Q) plus the work done on the system (W):
$\Delta U = Q+W$
In this example, no work is done on the bottle so W=0, while the heat given to the system is $Q=2.0 \cdot 10^4 J$ , so the change in internal energy of the gas is
$\Delta U = Q = 2.0 \cdot 10^4 J$

4 0

3 years ago

A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find

nikitadnepr [17]

Answer:

a) x = v₀ₓ t , y = $v_{oy}$ t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis

-W = m $a_{y}$

-m g = m $a_{y}$

$a_{y}$ = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

x = v₀ₓ t + ½ a t²

As the acceleration on this axis is zero

x = v₀ₓ t

Y Axis

y = $v_{oy}$ t + ½ $a_{y}$ t²

The acceleration on this axis is –g

y = $v_{oy}$ t - ½ g t²

B) to find the maximum value of distance r

r =√ x² + y²

r = √( v₀ₓ² t² + ( $v_{oy}$ t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

dr / dt = 0

0 = ½ 1/√( v₀ₓ² t² + ( $v_{oy}$ t + ½ g t²)² (v₀ₓ² 2t + 2 ( $v_{oy}$ t - ½ g t²)( $v_{oy}$ - ½ g2t)

We simplify this expression

0 = v₀ₓ² 2t + 2 ( $v_{oy}$ t - ½ g t²) ( $v_{oy}$ - ½ g2t)

-v₀ₓ² t =( $v_{oy}$ t - ½ g t²) ( $v_{oy}$ - ½ g2t)

-v₀ₓ² t = $v_{oy}$ ² t -3/2 gt² $v_{oy}$ + 1/2 g² t³

½ g² t²- 3/2 g $v_{oy}$ t = $v_{oy}$ ² + v₀ₓ²

Let's use trigonometry to find go and vox

sin θ = $v_{oy}$ / v₀

cos θ = vox / v₀

$v_{oy}$ = v₀ sin θ

v₀ₓ = v₀ cos θ

We replace

½ g² t² -3/2 g $v_{oy}$ t = v₀ (sin² θ + cos² θ)

g t² - 3v₀ sin θ t = 2 v₀/g

The time is maximum for the angle is zero

8 0

3 years ago

Key Stage 3 Science - Physics

stira [4]

Answer:

F = 3750 N

Explanation:

Given that,

Pressure, P = 150 Pa

Area, a = 25m²

We need to find the force applied. We know that, pressure is equal to the force acting per unit area. It can be given by :

$P=\dfrac{F}{A}\\\\F=P\times A\\\\F=150\ Pa\times 25\ m^2\\\\F=3750\ N$

So, the required force is 3750 N.

5 0

3 years ago

The position coordinate of a particle which is confined to move along a straight line is given by s =2t3−24t+6, where s is measu

Gre4nikov [31]

Answer:

a) the answer is t=4 seconds

b) acceleration is zero

c) displacement= 142 m

Explanation:

Given the position of the particle

$s=2t^3-24t+6$

a) the time required when velocity $v=72 m/s$

$v=72=\frac{ds}{st}=6t^2-24$

now we solve for time $t$

$6t^2=72+24=96\\\\\implies t^2=\frac{96}{6}\\\therefore t=4 s$

b) acceleration when $v=30 m/s$

acceleration is the time derivative of velocity i.e

$a=\frac{dv}{dt}=\frac{d}{dt}(30)=0$

c) the net displacement of the particle during the interval t = 1 s to t = 4 s is

$s_4-s_1=2t^3-24+6|_{t=4}-2t^3-24+6|_{t=1}\\s_4-s_1=126-(-16)=142 m$

7 0

4 years ago