Answer:
The acceleration of the box is 3 m/s²
Explanation:
Given;
mass of the box, m = 12 kg
horizontal force pulling the box forward, Fx = 48 N
frictional force acting against the box in opposite direction, Fk = 12 N
The net horizontal force on the box, F = 48 N - 12 N
The net horizontal force on the box, F = 36 N
Apply Newton's second law of motion to determine the acceleration of the box;
F = ma
where;
F is the net horizontal force on the box
a is the acceleration of the box
a = F / m
a = 36 / 12
a = 3 m/s²
Therefore, the acceleration of the box is 3 m/s²
Answer:
0.7 kg m²
Explanation:
F = force exerted applied by muscle in a professional boxer = 2551 N
r = length of lever arm = 3.15 cm = 0.0315 m
α = angular acceleration of the forearm = 115 rad/s²
I = moment of inertia of the boxer's forearm
τ = Torque applied by muscle in a professional boxer
Torque is given as
τ = I α = r F
Inserting the values
I (115) = (0.0315) (2551)
I = 0.7 kg m²
Answer:
158/25
Explanation:
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Answer:
300 N/m
Explanation:
given,
Load attached to the spring, W = 54 N
length of stretch of the spring, x = 0.15 m
spring constant= ?
Force applied on the spring is calculated by the equation
F = k x
where k is the spring constant
x is the displacement of the spring due to applied load
now,
54 = k × 0.15


hence, the spring constant is equal to 300 N/m
Explanation:
using the formula: S=ut+½gt², where u=0, S=?, g=8m/s², t=10seconds.
S=ut+½gt² ("ut" term will cancel because u=0).
=> S= ½gt²
=>S = ½×8×10²
=>S = 4×100
=>S = 400m .
Therefore, the distance traveled by the body in 10s is 400m.
hope this helps you.