Known :
l = 7 cm
w = 4 cm
Asked :
h = ...?
Answer :
V = B triangle × h (long)
35 = ½ × 4 × h × 7
35 = ½ × 28 × h
35 = 14 h
h = 35 ÷ 14
h = 2,5 cm
Sorry if I am wrong, I only study
The stake, height and tether length of the tent form a right angle triangle where the tether length is the hypotenuse.
Applying Pythagoras theorem:
length² = height² + (stake distance)²
length = √(8² + 2²)
length = 8.5 feet
Answer:
ax = -3.29[m/s²]
ay = -1.9[m/s²]
Explanation:
We must remember that acceleration is a vector and therefore has magnitude and direction.
In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.
![a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]](https://tex.z-dn.net/?f=a_%7Bx%7D%3D-3.8%2Acos%2830%29%20%3D%20-3.29%20%5Bm%2Fs%5E%7B2%7D%5D%5C%5C%20a_%7By%7D%3D-3.8%2Asin%2830%29%20%3D%20-1.9%20%5Bm%2Fs%5E%7B2%7D%5D)
A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:
F=K(q1xq2)/r²
F: The magnitud of the force between the charges. (F=2.0 N).
K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
q1 and q2: Electrical charges.
r: The distance between the charges (r=1.35 m).
We have the values of F, K and r, so we can calculate q1xq2, because both<span> coins have identical charges:
</span>
q1xq2=(r²xF)/K
q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
q1xq2=3x10^-10 C
q1=q2=(<span>3x10^-10 C)/2
</span>Then, the charge of each coin, is:
<span>
q1=1.5x</span><span>10^-10 C
</span>q2=1.5x10^-10 C
B) <span>Would the force be classified as a force of attraction or repulsion?
</span>
It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.
