Answer:
a) fr = 224.3 N
, b) fr = 224.3 N
, c) v = 198.0 m/s
Explanation:
a) For this exercise let's start by calculating the acceleration in the fall
v² = v₀² - 2 a (y-y₀)
When it jumps the initial vertical speed is zero
a = -v² / 2 (y-y₀)
a = -68 2/2 (1000-2000)
a = 2,312 m / s²
Let's use the second net law to enter the average friction force
fr = m a
fr = 97 2,312
fr = 224.3 N
b) let's look for acceleration
v² = v₀² - 2 a y
a = (v² –v₀²) / 2 (y-y₀)
a = (4² - 68²) / 2 (0-1000)
a = 2,304 m / s²
fr = m a
fr = 97 2,304
fr = 223.5 N
c) the speed of the wallet is searched with kinematics
v² = v₀² - 2 g (y-y₀)
v = √ (0-2 9.8 (0-2000))
v = 198.0 m/s
Answer: 20.2 m/s
Explanation:
From the question above, we have the following data;
M1 = 800kg
M2 = 1200kg
V1 = 13m/s
V2 = 25m/s
U (common velocity) =?
M1V1 + M2V2 = (M1 + M2). U
(800*13) + (1200*25) = (800+1200) * U
10400 + 30000 = 2000u
40400 = 2000u
U = 40400 / 2000
U = 20.2 m/s
ANSWER

EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:

That is the answer.
It's highly reactive and contains only one valence electron