Answer:
Period of one vibration = 0.00215 second (Approx.)
Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)
Explanation:
Given:
Frequency of wave = 466 Hz
Find:
Period of one vibration
Wavelength {Is speed of sound is 343 m/s]
Computation:
Period of one vibration = 1/F
Period of one vibration = 1 / 466
Period of one vibration = 0.00215 second (Approx.)
Wavelength = Velocity / Frequency
Wavelength {Is speed of sound is 343 m/s] = 343 / 466
Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)
Answer:
1.5 unit of product per min
Explanation:
30 units of product was moved in 30 minutes.
Number of units left = Total number of units-number of units moved
=75-30 =45 units
45 units is available to be moved for the rest 30 min. To be able to achieve this goal of 75 units of product per hour.
45/30 amount of units must be moved in 1 min
=1.5 unit per min
Question:
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
(a)60 (b)90 (c)120
Answer:
(a)5.42 N (b)6.26 N (c)5.42 N
Explanation:
From the question
Length of wire (L) = 2.80 m
Current in wire (I) = 5.20 A
Magnetic field (B) = 0.430 T
Angle are different in each part.
The magnetic force is given by

So from data

Now sub parts
(a)

(b)

(c)
