Answer:
a) equal 1, b) less than 1
Explanation:
a) the electric force is given by
Fe = q E
The charge of the electron and proton has the same value, that of the proton is positive and that of the electron is negative
Proton
Fp = qE
Electron
Fe = - q E
Fp / Fe = -1
If we do not take into account the sign the relationship is equal to one (1)
b) to calculate the force we use Newton's second law
F = ma
qE = m a
a = q E / m
The mass of the proton much greater than the mass of the electron
ap = q E / 
ae = - q E / 
ap / ae = /
= /1600 =1/1600
It is much smaller than 1
Answer:
a) dh/dt = -44.56*10⁻⁴ cm/s
b) dr/dt = -17.82*10⁻⁴ cm/s
Explanation:
Given:
Q = dV/dt = -35 cm³/s
R = 1.00 m
H = 2.50 m
if h = 125 cm
a) dh/dt = ?
b) dr/dt = ?
We know that
V = π*r²*h/3
and
tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5
⇒ h = (5/2)*r
⇒ r = (2/5)*h
If we apply
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π
a) if r = (2/5)*h
⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π
⇒ (4/25)(3*h²)(dh/dt) = -105/π
⇒ dh/dt = -875/(4π*h²)
b) if h = (5/2)*r
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π
⇒ (5/2)*(3*r²)(dr/dt) = -105/π
⇒ dr/dt = -14/(π*r²)
Now, using h = 125 cm
dh/dt = -875/(4π*h²) = -875/(4π*(125)²)
⇒ dh/dt = -44.56*10⁻⁴ cm/s
then
h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)
⇒ r = 50 cm
⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)
⇒ dr/dt = -17.82*10⁻⁴ cm/s
Answer:
x₁ = 58.09 m
Explanation:
Displacement is calculated by finding the final distance away from a point then subtracting the initial distance.
Given that initial position of object is x=25.89 m
Displacement Δx=32.2 m
Final position x₁=?
Δx = x₁-x
32.2= x₁-25.89
32.2+25.89 = x₁
58.09 m =x₁
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass of the other block is 
Explanation:
From the question we are told that
Mass of the first block is 
The height is 
The time it takes it is 
Generally from kinematic equation
Here u is the initial velocity which zero given that it was at rest initially
So
=> 
=> 
=> 
Generally from the diagram the resultant force due to the weight of the first object and the tension on the string is mathematically represented as
=> 
=> 
=> 
Generally from the diagram the resultant force due to the weight of the second object and the tension on the string is mathematically represented as
=> 
=> 
=>
