Answer:
x = 1474.9 [m]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.
We must understand that when forces are applied on the body, they tend to slow the body down to stop it.
So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.
ΣF = m*a
![10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]](https://tex.z-dn.net/?f=10%20%2B12%2Asin%2860%29%3D%20-%206%2Aa%5C%5Ca%20%3D%20-%203.39%5Bm%2Fs%5E%7B2%7D%5D)
Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].
where:
Vf = final velocity = 0 (the block stops)
Vo = initial velocity = 100 [m/s]
a = - 3.39 [m/s²]
x = displacement [m]
![0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]](https://tex.z-dn.net/?f=0%20%3D%20100%5E%7B2%7D-2%2A3.39%2Ax%5C%5Cx%3D%5Cfrac%7B10000%7D%7B2%2A3.39%7D%5C%5Cx%3D1474.9%5Bm%5D)
Answer:
8.46E+1
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 39 C
Charge 2 (q₂) = –53 C
Force (F) of attraction = 26×10⁸ N
Electrical constant K) = 9×10⁹ Nm²/C²
Distance apart (r) =?
The distance between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
26×10⁸ = 9×10⁹ × 39 × 53 / r²
26×10⁸ = 1.8603×10¹³ / r²
Cross multiply
26×10⁸ × r² = 1.8603×10¹³
Divide both side by 26×10⁸
r² = 1.8603×10¹³ / 26×10⁸
r² = 7155
Take the square root of both side
r = √7155
r = 84.6 m
r = 8.46E+1 m
Answer:
h = 23.716 m
Explanation:
Given that,
The time taken by the stone to hit the water is, t = 2.2 s
Height of the bridge above the ground, h = ?
The distance that the body will fall through the time is given by the formula
S = 1/2 gt² m
Where,
g - acceleration due to gravity
Substituting the values in the above equation
S = 1/2 x 9.8 m/s² x (2.2 s)²
= 23.716 m
Therefore, the height of the bridge from the surface of the water is h = 23.716 m
During the first phase of acceleration we have:
v o = 4 m/s; t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?; v o = 13 m/s; a = 1.125 m/s² ; t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s
v^2 = v0^2 +2ad
v^2 = 22^2 + 2*3.78*45 = 824.2
v= √824.2 = 28.7 m/s
