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Morgarella [4.7K]
3 years ago
8

What is the net charge of a copper atom if it gains 2 electrons?

Physics
1 answer:
Alex_Xolod [135]3 years ago
7 0
If an atom gains electrons, it develops a negative charge equal to the number of electrons gained.
So the net charge on the copper atom which gained 2 electrons will be -2.
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Kuiper Belt objects are composed of which substances? Select all that apply.
levacccp [35]
IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron

8 0
3 years ago
A metabolic waste of algae that can be recycled for use in cellular respiration is
m_a_m_a [10]

Answer:oxygen

Explanation:algae are plants that lives in aquatic habitat,with a few of them occurring in land. Algae have chlorophyll and as a result are autotrophic in nutrition. Algae uses carbon dioxide as a raw material for photosynthesis which is the process where they produce food. They also give off oxygen which results from the splitting of water by light.

This oxygen given off is used by organisms for cellular respiration.the mitochondria is the organelle responsible it's utilization in respiration and carbon dioxide is given off.oxygen serves as an electron acceptor in the energy producing process in the mitochondria. It is an important gas for aerobic respiration.

5 0
3 years ago
In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8
BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

      F = m a

      a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

       P = I / c        absorbent surface

       P = F / A

       F / A = I / c

       F = I A / c

The area of ​​area of ​​a circle is

      A = π r²

We replace

     F = I π r² / c

Let's calculate

     F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

     F = 8.375 10⁻²³ N

Density is

      ρ = m / V

      m = ρ V

      m = ρ (4/3 π r³)

      m = 4500 (4/3 π (1 10⁻⁶)³)

      m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

     a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

     a = 4.44 10⁻⁹ m/s²               absorbent surface

The correct answer is B

4 0
3 years ago
In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
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