Explanation: (I think)
Plug your values into the momentum equation.
So m1= 63kg
m2 = 10 kg
V1 = 12 m/s
And then plug in your values and solve for your unknown (v2)
Answer:
speed wind Vw = 54.04 km / h θ = 87.9º
Explanation:
We have a speed vector composition exercise
In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south
Let's add the vectors in each coordinate axis
X axis (East-West)
-Xvion - Xw = -119
Xw = -Xavion + 119
Xw = 119 -108
Xwi = 1 km
Calculate the speed for time of t = 0.5 h
Vwx = Xw / t
Vwx= 1 /0.5
Vwx = - 2 km / h
Y Axis (North-South)
Y plane - Yi = -27
Y plane = 0
Yw = 27 km
Vwy = 27 /0.5
Vwy = 54 km / h
Let's use the Pythagorean theorem and trigonometry to compose the answer
Vw = √ (Vwx² + Vwy²)
Vw = R 2² + 54²
Vw = 54.04 km / h
tan θ = Vwy / Vwx
tan θ = 54/2 = 27
θ = Tan⁻¹ 1 27
θ = 87.9º
The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º west from the south
The wind Wind turbines work on a simple principle: instead of using electricity to make wind—like a fan—wind turbines use wind to make electricity. Wind turns the propeller-like blades of a turbine around a rotor, which spins a generator, which creates electricity.
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
Explanation:
Image distance, v = -17 cm (-ve for virtual image)
Radius of curvature of concave mirror, R = 39 cm
Focal length, f = -19.5 cm (-ve for a concave mirror)
(a) Using mirror's formula as :


u = 132.6 cm
So, the object is placed 132.6 cm in front of the mirror.
(b) Magnification of the mirror, 

m = -0.128
Hence, this is the required solution.