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3241004551 [841]
4 years ago
15

The element copper has 29 protons. What is the atomic number of copper?

Physics
2 answers:
Mumz [18]4 years ago
7 0
29.....the number of protons is also the same as the atomic number.
Aleksandr-060686 [28]4 years ago
3 0

Answer:

29

Explanation:

We are given that

An element has 29 protons.

Number of protons=29

We have to find the atomic number of copper.

We know that

Atomic number of an element:

Atomic number is defined as the number of protons of an  element.

It is represented by Z.

Number of protons=Atomic number =Z

Therefore, Atomic number of copper=Z=29

Answer: 29

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Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less thanthat o
snow_tiger [21]

Answer:

41.4 g

41.4 cm³

1.08695 g/cm³

Explanation:

\rho = Density of water = 1 g/cm³

Mass of water displaced will be the difference of the

m=45-3.6\\\Rightarrow m=41.4\ g

Mass of water displaced is 41.4 g

Density is given by

\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{41.4}{1}\\\Rightarrow v=41.4\ cm^3

So, volume of bone is 41.4 cm³

Average density of the bird is given by

\rho=\dfrac{45}{41.4}\\\Rightarrow \rho=1.08695\ g/cm^3

The average density is 1.08695 g/cm³

3 0
3 years ago
.
WARRIOR [948]

Answer:

Explanation:distance-time =speed

a,V =s/t

V=2m/2s

V=1m/s

b,v=s/t

V=80m/40s

V= 2m/s

The average speed is 2m/s

8 0
2 years ago
A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has
LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

<h3>What is the distance?</h3>
  • The separation of one thing from another in space; the distance or separation in space between two objects, points, lines, etc.; remoteness. The distance of seven miles cannot be accomplished in one hour of walking.
  • Learn how to use the Pythagorean theorem to get the separation between two points using the distance formula. The Pythagorean theorem can be rewritten as d==(((x 2-x 1)2+(y 2-y 1)2)
  • The distance between any two places is the length of the line segment separating them. By measuring the length of the line segment that connects the two points in coordinate geometry, the distance between them may be calculated.

(c) the distance D where the object comes to rest.

W_{total} =ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1*2.3^{2}

⇒D=\frac{0.5*2.3^{2} }{2.45}

⇒1.08.m

To learn more about distance, refer to:

brainly.com/question/4998732

#SPJ4

5 0
2 years ago
Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we
sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

<u>Given the following data:</u>

  • Mass 1 = 80 kg
  • Velocity 1 = 2 m/s east.
  • Mass 2 = 70 kg
  • Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

Momentum = mass \times velocity

<u>For Football player 1:</u>

Momentum = 80 \times 2

Momentum 1 = 160 Kgm/s.

<u>For Football player 2:</u>

Momentum = 70 \times 3

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160

Total momentum = 50 Kgm/s.

<u>Note:</u> We subtracted because the football players were moving in opposite directions.

Read more: brainly.com/question/15517471

6 0
3 years ago
astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o
devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ T=2 \pi\sqrt{\frac{m}{k} }

or,

⇒ m=\frac{k T^2}{4 \pi^2}

On substituting the values, we get

⇒     =\frac{280\times (3)^2}{4 \pi^2}

⇒     =\frac{280\times 9}{4\times (3.14)^2}

⇒     =68.83 \ kg

(b)

The speed of the string will be:

⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

then,

⇒             v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }

On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

⇒                =0.725 \ m/s

4 0
3 years ago
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