Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
The magnitude of the electric force on a protein with this charge is
Explanation:
Given that,
Electric field = 1500 N/C
Charge = 30 e
We need to calculate the magnitude of the electric force on a protein with this charge
Using formula of electrostatic force
Where, F = force
E = electric field
q = charge
Put the value into the formula
Hence, The magnitude of the electric force on a protein with this charge is