<em>weight = (mass) x (gravity)</em>
Weight = (5.00 kg) x (9.81 m/s²)
weight = (5.00 x 9.81) (kg-m/s²)
<em>Weight = 49.05 Newton</em>
Answer:
27 blocks
Explanation:
First, the expression to use here is the following:
P = F/A
Where:
P: pressure
F: Force exerted
A: Area of the block.
Now , we need to know the number of blocks needed to exert a pressure that equals at least 2 atm. To know this, we should rewrite the equation. We know that certain number of blocks, with the same weight and dimensions are putting one after one over the first block, so we can say that:
P = W/A
P = n * W1 / A
n would be the number of blocks, and W1 the weight of the block.We have all the data, and we need to calculate the area of the block which is:
A = 0.2 * 0.1 = 0.02 m²
Solving now for n:
n = P * A / W1
The pressure has to be expressed in N/m²
P = 2 atm * 1.01x10^5 N/m² atm = 2.02x10^5 N/m²
Finally, replacing all data we have:
n = 2.02x10^5 * 0.02 / 150
n = 26.93
We can round this result to 27. So the minimum number of blocks is 27.
Answer:
Its momentum is multiplied by a factor of 1.25
Explanation:
First, we <u>calculate the initial velocity of the object</u>:
- 59.177 J = 0.5 * 3.4 kg * v₁²
With that velocity we can <u>calculate the initial momentum of the object</u>:
Then we <u>calculate the velocity of the object once its kinetic energy has increased</u>:
- (59.177 J) * 1.57 = 0.5 * 3.4 kg * v₂²
And <u>calculate the second momentum of the object</u>:
Finally we <u>calculate the factor</u>:
Pe = mgh.
14000 J = (40kg)(10m/s^2)(h)
h = 35 meters
Answer:
E = 
Explanation:
For this exercise let's use Gauss's law. The Gaussian surface that follows the symmetry of the charges is a sphere
Ф = ∫ E. dA = 
the bold are vectors, the radii of the sphere and the electric field are parallel therefore the scalar product reduces to the algebraic product
Ф = ∫ E dA = \frac{x_{int} }{\epsilon_o}
E ∫ dA = \frac{x_{int} }{\epsilon_o}
E A = \frac{x_{int} }{\epsilon_o}
the area of a sphere is
A = 4π r²
the charge inside the sphere is q = + q
we substitute
E 4π r² = \frac{x }{\epsilon_o}
E =
