Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
Average speed = total distance traveled/total time taken
it is the total distance traveled in a total time the total distance is attained
Answer:
Explanation:
Project mass m=3.8 kg
Initial speed vi= 0m/s
Final speed vf= 9.3×10³ m/s
Force F=9.3×10⁵N
To find
Time t
Solution
From Newtons second law we know that
∑F=ma
Where m is mass
a is acceleration
We can write this equation as
∑F=m(Δv/Δt)
Rearrange this equation to find time t
So
Substitute the given values
Answer:
a. The thickness of the wire is 2.5 mm.
b. The wire is 0.25 cm thick.
Explanation:
Number of turns of the wire = 10
The length of total turns = 25 mm
a. The thickness of the wire can be determined by;
thickness of the wire =
=
= 2.5 mm
Therefore, the wire is 2.5 mm thick.
b. To determine the thickness of the wire in centimetre;
10 mm = 1 cm
So that,
2.5 mm = x
x =
= 0.25 cm
The wire is 0.25 cm thick.
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N