Answer:
Explanation:
The question is incomplete.
The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.
s= sin2(pi)t
Acceleration = d²S/dt²
dS/dt = 2πcos2πt
d²S/dt² = -4π²sin2πt
A(t) = -4π²sin2πt
Next is to find acceleration after 4.5 seconds
A(4.5) = -4π²sin2π(4.5)
A(4.5) = -4π²sin9π
A(4.5) = -4π²sin1620
A(4.5) = -4π²(0)
A(4.5) = 0m/s²
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
Answer:
0.037 A
Explanation:
Magnetic field = B = 1.00 e-4 T
Length = L = 0.380 m
Number of turns = 810
B = μ₀ N I / L
⇒ Current = I = B L / μ₀ N = ( 1 e-4) ( 0.380) / (4π × 10⁻⁷)(810)
= 0.037 A = 37.3 mA
An external force that is being applied in the direction of the displacement
Answer:
70 cm
Explanation:
0.5 kg at 20 cm
0.3 kg at 60 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)
The position of the third mass of 0.6 kg is at 20+50 = 70 cm