Answer:
ok so
Explanation:
Im not sure rn but ill get back to you.
The acceleration is 3.3 m/s2
Answer:
a) = 10.22 rad/s
b) = 0.35 m
Explanation:
Given
Mass of the particle, m = 1.1 kg
Force constant of the spring, k = 115 N/m
Distance at which the mass is released, d = 0.35 m
According to the differential equation of s Simple Harmonic Motion,
ω² = k / m, where
ω = angular frequency in rad/s
k = force constant in N/m
m = mass in kg
So,
ω² = 115 / 1.1
ω² = 104.55
ω = √104.55
ω = 10.22 rad/s
If y(0) = -0.35 m and we want our A to be positive, then suffice to say,
The value of coefficient A in meters is 0.35 m
Answer:
Explanation:
The magnetic field produced by a current-carrying wire is given by
where
is the vacuum permeability
I is the current
r is the distance from the wire
In this problem we have
r = 1.2 mm = 0.0012 m
So the magnetic field strength is
Answer:
159.38 Watts
Explanation:
Initially;
- Mass on the spring is 8.5 kg
- Therefore, compression force is 85 N
- Compression distance is 15 cm or 0.15 m
But;
F = kx
where F is the force of compression, k is the spring constant and x is the compression distance.
Thus;
k = F/x
= 85 N ÷0.15
= 566.67 N/m
We are required to determine the power needed to stretch the same spring for 1.5 m in 4 secs.
Power = Work done ÷ time
Work done is given by 0.5kx²
Therefore;
Power = 0.5kx²÷ t
= (0.5×566.67 N/m × 1.5² ) ÷ 4 seconds
= 159.38 Watts
Thus, the power needed is 159.38 watts
