The formula we can use in this case is:
d = v0t + 0.5 at^2
v = at + v0
where,
d = distance travelled
v0 = initial velocity = 0 since at rest
t = time travelled
a = acceleration
v = final velocity when it took off
a. d = 0 + 0.5 * 3 * 30^2
d = 1350 m
b. v = 3 * 30 + 0
<span>v = 90 m/s</span>
Answer:
True
Explanation:
Because I had a test on this
Answer:
Explanation:
The magnitude of the electrostatic force between two charged objects is
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.
In this problem, we have:
is the distance between the charges
since the charges are identical
is the force between the charges
Re-arranging the equation and solving for q, we find the charge on each drop:
The hot sun beating down on the grass is hot because the grass aborbs the heat from the sun
Merry Christmas!!!
-Kaden&Sydney
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
