Answer:
Explanation:
a) Using the equation of motion
S = ut + 1/2gt²
S is the distance of fall
g is the acceleration due to gravity
t is the time taken
Given S = 12.0m, g = 9.81m/s^2, un= 0m/s
12 = 0+1/2(9.81)t²
12 = 4.905t²²²
t² = 12/4.905
t² = 2.446
t = √2.446
t = 1.56secs
b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt
v = 0+9.81(1.56)
v = 15.34m/s
Hence the frog is falling at the rate of 15.34m/s
Answer:
15
Explanation:
P=W/T
T=6sec
W=?
F=60N
S=18m
W=F X S. .s indicate displacement
W=60x18
W=108
So p=108 j/6sec
P=15watt
Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
(D)
to establish an understanding of key concepts relating to population biology
Explanation:
Thats what I would go with but I didn't read the article so I don't know what context was used. Good luck! :)
Answer:
The coefficient is 0.90
Explanation:
Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.
