Answer:
Explanation:
Volume of block A = 10 x 6 x 1 = 60 cm³
Mass of block A = 630 g
density of mass A = mass / density
= 630 / 60 = 10.5g / cm³
Volume of block B = 5 x 5 x 3 = 75 cm³
Mass of block A = 604 g
density of mass A = mass / density
= 604 / 75 = 8.05 g / cm³
Since density of both A and B are less than that of mercury , both will float in mercury.
Answer:
Intensity of beam 18 feet below the surface is about 0.02%
Explanation:
Using Lambert's law
Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium
then dI / I = kdt
taking log,
ln(I) = kt + ln C
I = Ce^kt
t=0=>I=I(0)=>C=I(0)
I = I(0)e^kt
t=3 & I=0.25I(0)=>0.25=e^3k
k = ln(0.25)/3
k = -1.386/3
k = -0.4621
I = I(0)e^(-0.4621t)
I(18) = I(0)e^(-0.4621*18)
I(18) = 0.00024413I(0)
Intensity of beam 18 feet below the surface is about 0.2%
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Answer: 10 and 35 degrees
Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment
Incomplete question.The complete question is here
What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.
Answer:
F= 0.034 N
Explanation:
Given Data
Outer=2 cm
Edge of blade=21 cm
Mass=7.7 g
Length of blade=21 cm
The last 2cm is treated as if they were at a point 20cm from the center of rotation
To Find
Force=?
Solution
Convert the given frequency to angular frequency
ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)
ω= 3/2*π rad/sec
Now to find centripetal force.
F = m×v²/r
F= m×ω²×r
Put the data
F = 0.0077 kg × (3/2×π rad/sec
)²× 0.20 m
F= 0.034 N
