Detergents are special, powerful cleansers that can break up dirt, oils, and grease in clothing or on dishes.
Cleaning solvents are used to remove oil, grease, solder flux, and other contaminants.
Acid cleaners are generally used to remove mineral deposits and are useful for descaling dishwashers or removing rust from restroom facilities.
Abrasive uses
* Buffing.
* Honing.
* Drilling.
* Grinding.
* Sanding.
* Polishing.
* Cutting.
* Sharpening.
Answer:
Explanation:
We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .
f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .
f will be lowest when v₀ is highest .
velocity of observer is highest when he is at the equilibrium position or at middle point .
So apparent frequency is lowest when observer is at the middle point and going away from the source while swinging to and from before the source of sound .
Answer:
2550kgm/s
Explanation:
Angular momentum is given as
P=mv
And V=w²r
W=25rad/s
r=34cm approx. 0.34m
V= 25²X 0.34
V=212.5m/s
To find the angular momentum,
P=mv
P= 12 X 212.5
P=2550kgm/s
Answer:
C = 2.9 10⁻⁵ F = 29 μF
Explanation:
In this exercise we must use that the voltage is
V = i X
i = V/X
where X is the impedance of the system
in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is
X =
tells us to take inductance L = 0.
The angular velocity is
w = 2π f
the current is required to be half the current at high frequency.
Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small
→0 when w → ∞
therefore in this frequency regime
X₀ =
the very small fraction for which we can despise it
X₀ = R
to halve the current at f = 200 H, from equation 1 we obtain
X = 2X₀
let's write the two equations of inductance
X₀ = R w → ∞
X= 2X₀ = w = 2π 200
we solve the system
2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }
4 R² = R² + 1 / (wC) ²
1 / (wC) ² = 3 R²
w C =
C =
let's calculate
C =
C = 2.9 10⁻⁵ F
C = 29 μF
Answer:
Group 1A (or IA) of the periodic table are the alkali metals: hydrogen (H), lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr).
Explanation:
Hope it helps! Correct me if I am wrong :>