According to the net force, the acceleration of the book is 16.47 m/s².
We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as
∑F = m . a
where F is force, m is mass and a is acceleration
From the question above, we know that
m = 3 kg
g = 9.8 m/s²
F1 = 20 N
Find the net force
∑F = F1 + W
∑F = 20 + m . g
∑F = 20 + 3 . 9.8
∑F = 20 + 29.4
∑F = 49.4 N
Find the acceleration
∑F = m . a
49.4 = 3 . a
a = 16.47 m/s²
Find more on force at: brainly.com/question/25239010
#SPJ4
The human body is connected in every way. All the organs are connected and help each other be alive. For example, the veins are connected to the heart, which help it by pumping blood and oxygen. If they weren’t there, the heart wouldn’t be able to sustain a life.
I really hope this gave you and ideas and helped you in some way:)
velocity of the physics instructor with respect to bus

acceleration of the bus is given as

acceleration of instructor with respect to bus is given as

now the maximum distance that instructor will move with respect to bus is given as




so the position of the instructor with respect to door is exceed by

so it will be moved maximum by 3 m distance
Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Given that,
Radius of track, r = 50 m
time , t = 9 s
velocity, v = ?
Distance covered by car in one lap around a track is equal to the circumference of the track.
C = 2 π r = 2 * 3.14 * 50
C = 314.159 m
Distance covered by car, s = 314.159 m
Velocity = distance/ time
V = 314.159 / 9
V = 34.9 m/s
The average velocity of car is 34.9 m/s.