Answer:
3.5m/s^2
Explanation:
From Newton's second Law of Motion
F = ma
Where F is the applied force, m is the mass of the object and a is the acceleration.
F = 350 N
Mass = 100kg
350N = 100×a
a = 350/100
a = 3.5m/s^2
The acceleration of the object will be 3.5m/s^2
Answer:
8 KJ/ s
Explanation:
Heat pumps Transfer thermal energy through absorbing of heat that comes from cold region and then release to warmer area by utilizing external power.
The coefficient of performance known as COP provide the ratio of both heating and cooling that are supplied to required work.
✓QH=The rate at which heat is produced = ?
✓COP= Coefficient of performance of a residential heat pump = 1.6
✓ W(in)= power consumption= 5KW
QH=The rate at which heat is produced=[Coefficient of performance of a residential heat pump] × [power consumption]
= 1.6 × 5KW
=8 KJ/ s
Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
It's a multiple of 5 because you can't multiply anything by 2 and get 25
