<u>Given that:</u>
Ball dropped from a bridge at the rate of 3 seconds
Determine the height of fall (S) = ?
As we know that, S = ut + 1/2 ×a.t²
u =initial velocity = 0
a= g =9.81 m/s (since free fall)
S = 0+ 1/2 × 9.81 × 3²
<em> S = 44.145 m</em>
<em>44.145 m far is the bridge from water</em>
It was in Texas on September 8, 1900.
Answer:
m = 236212 [kg]
Explanation:
The potential energy can be determined by means of the product of mass by gravity by height. In this way, we have the following equation.

where:
P = potential energy = 3360000000 [J]
m = mass [kg]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 1450 [m]
Now, we can clear the mass from the equation above:
![3360000000=m*9.81*1450\\m = 236212 [kg]](https://tex.z-dn.net/?f=3360000000%3Dm%2A9.81%2A1450%5C%5Cm%20%3D%20236212%20%5Bkg%5D)
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:

The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) =
₀
V(r) =
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:

= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to