The desire for positive reinforcement.
Answer: 0.642mm
Explanation: F= force = 5.2×10^-16 N,
v = velocity of electron = 1.2×10^7 m/s,
m = mass of electron = 9.11×10^-31 kg.
We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.
Recall that f = ma.
Where a = acceleration
This acceleration of vertical because it occurred when the object deflected.
5.2×10^-16 = 9.11×10^-31 (ay)
ay = 5.2×10^-16 / 9.11×10^-31
ay = 5.71×10^14 m/s²
For the horizontal motion, x = vt
Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,
By substituting the parameters, we have that
0.019 = 1.27×10^7 × t
t = 0.019 / 1.27 × 10^7
t = 1.5×10^-9 s
The vertical distance (y) is gotten by using the formulae below
y = ut + at²/2
but u = 0
y = at²/2
y = 5.71×10^14 × (1.5×10^-9)²/2
y = 0.00128475/2
y = 0.000642m = 0.642mm
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
Answer:
93 km/h
Explanation:
Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h
Let the first 5 hours journey distance = F
From the formula of speed,
Speed = distance/time
Substitute speed and time
72 = F/5
F = 72 × 5 = 360 km
The remaining distance will be:
639 - 360 = 279km
The remaining time will be:
8 - 5 = 3 hours
Speed = 279/3
Speed = 93 km/h
Therefore, the average speed for the remaining time of the journey is equal to 93 km/h
AnMolar mass of CuCO3 = 123.5549 g/mol
This compound is also known as Copper(II) Carbonate.
Convert grams CuCO3 to moles or moles CuCO3 to grams
Molecular weight calculation:
63.546 + 12.0107 + 15.9994*3
Percent composition by element
Element Symbol Atomic Mass # of Atoms Mass Percent
Copper Cu 63.546 1 51.431%
Carbon C 12.0107 1 9.721%
Oxygen O 15.9994 3 38.848%
Explanation: