Answer:
The answer is 4200 J.
Explanation:
The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :
W = F×D
F = 42N
D = 100m
W = 42 × 100
= 4200 J
Answer:
Explanation:
1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.
Answer:
Explanation:
Position of charge 3q is x = 0
position of charge -2q is x = a
so here we know that
when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge
So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q
so here we can say
so we will have
so the x coordinate of this position is given as