Scott-Dannemiller, Koeninger, Briscoe, and Carter
Christopher-Briscoe, Dannemiller, and Koeninger
Dianne-Koeninger, Briscoe, and Carter
Kailee-Koeninger and Carter
Answer:
Explanation:
Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.
Therefore,
u = 10 m/s, initial upward velocity.
H = - 20 m, position of the ground.
g = 9.8 m/s², acceleration due to gravity.
Part (a)
When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore
u² - 2gh = 0
h = u²/(2g) = 10²/(2*9.8) = 5.102 m
At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.
Answer: 25.1 m above ground
Part (b)
Let v = the velocity when the frog hits the ground. Then
v² = u² - 2gH
v² = 10² - 2*9.8*(-20) = 492
v = 22.18 m/s
Answer: The frog hits the ground with a velocity of 22.2 m/s
Answer:
in 90 seconds It'll travel 450m At constant speed of 5m/s
I don’t know what it’s is
a) Southward you need to apply right hand rule. If you close your hand to the east, your thumb will indicate south.
b) Given the equation for Magnetic Force
Replacing
c) Given the second Newton's Law by
Given the electric force by,
