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Zolol [24]
3 years ago
14

Mass number of an atom with 11 protons

Physics
1 answer:
Ratling [72]3 years ago
8 0
The atomic mass of sodium (Na) is 22.98977
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A person is sitting at the very back of a canoe of length L, when the front just bumps into the dock. show answer No Attempt 50%
Pavel [41]

The distance of the canoeist from the dock is equal to length of the canoe, L.

<h3>Conservation of linear momentum</h3>

The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.

v(m₁ + m₂) = m₁v₁ + m₂v₂

where;

v is the velocity of the canoeist and the canoe when they are together

  • u₁ is the velocity of the canoe
  • u₂ velocity of the canoeist
  • m₁ mass of the canoe
  • m₂ mass of the canoeist

<h3>Distance traveled by the canoeist</h3>

The distance traveled by the canoeist from the back of the canoe to the front of the canoe is equal to the length of the canoe.

Thus, the distance of the canoeist from the dock is equal to length of the canoe, L.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0
2 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
How does the saturation of a solution affect crystal formation? In your own words please
ryzh [129]

The term saturated solution is used in chemistry to define a solution in which no more solute can be dissolved in the solvent. It is understood that saturation of the solution has been achieved when any additional substance that is added results in a solid precipitate or is let off as a gas.

3 0
3 years ago
Which of the following is the best reason for using a scanning electron microscope?
torisob [31]

Answer is C: Ability to see three-dimensional images of the surfaces of object

Explanation:

To enable the technician see fractures and broken particles in a better resolution as the SEM sees the peaks and valley of the structure.

8 0
2 years ago
Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitu
Elden [556K]

Answer:

Ex= -23.8 N/C  Ey = 74.3 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  88.6 N/C + (-14.3 N/C) =74.3 N/C

7 0
3 years ago
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