The answer is d.heuristic
The ball rolled a distance of
d = 12m + 20m.
But the change of position is
x = + 12m - 20m
Answer:
F_Balance = 46.6 N ,m' = 4,755 kg
Explanation:
In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in
∑ F = 0
Fe –W + F_Balance = 0
F_Balance = - Fe + W
The electric force is given by Coulomb's law
Fe = k q₁ q₂ / r₂
The weight is
W = mg
Let's replace
F_Balance = mg - k q₁q₂ / r₂
Let's reduce the magnitudes to the SI system
q₁ = + 8 μC = +8 10⁻⁶ C
q₂ = - 3 μC = - 3 10⁻⁶ C
r = 0.3 m = 0.3 m
Let's calculate
F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²
F_Balance = 49 - 2,397
F_Balance = 46.6 N
This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.
Mass reading is
m' = F_Balance / g
m' = 46.6 /9.8
m' = 4,755 kg
Answer:
<h2>82.94 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 28.8 × 2.88 = 82.944
We have the final answer as
<h3>82.94 N</h3>
Hope this helps you
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F
× t = m( v - v₀ )
we substitute
F
× 0.00265 = 1.75( 0 - 5.25 )
F
× 0.00265 = 1.75( - 5.25 )
F
× 0.00265 = -9.1875
F
= -9.1875 / 0.00265
F
= -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F
= - F
we substitute
F
= - ( -3466.98 N )
F
= 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N