Answer:
displacement= 30 m towards south, distance= 210m
Explanation:
Distance (scalar quantity) how much ground an object has covered.
Displacement (vector quantity) refers to how far out of place an object is it is the object's overall change in position.
Basically meaning for displacement the directions will be very key
D for Displacement
D= D1+D2
D= 120 (S) + 90 m (N)
Must be in same direction
D= 120 (S) + (-90 m) (S)
D= 30 m (S)
and for distance you are simply just adding how much distance they have covered
so d= d1+d2
d= 90m + 120m
d= 210m
Answer:
Magnitude = 4.056 m
Direction = 42.3⁰
Explanation:
The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.
The vector 4.40 is directed East. This automatically becomes a horizontal component.
But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.
Resolving the vectors should yield the horizontal and vertical components:
Horizontal components
The first component is 4.40 m
The second one is derived by resolving 3.40 to the horizontal like this 3.40 × - cos 61⁰ = -1.648 m
Adding the horizontal component gives 4.40 m + ( -1.648 m) = 2.752 m
Vertical components
Resolve 3.40 with the angle 61⁰ like this: vertical comp = 3.41 × sin 61
= 2.98 m
The magnitude is given by √[(2.98)²+ (2.752)²] = 4.056 m Ans
The direction us given by tan⁻¹ (2.98/2.752) = 42.3⁰ Ans
Answer:
t = 2.68 x 10¹⁴ years
Explanation:
First we need to find the amount of energy that Sun produce in one day.
Energy = Power * Time
Energy of Sun in 1 day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/ 1 hr)
Energy of Sun in 1 day = 3.32 x 10³¹ J
Now, the time required by the nuclear power generator, in years, will be:
Energy of power generator = Energy Sun in 1 day = 3.32 x 10³¹ J
3.32 x 10³¹ J = Power * Time
3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)
t = 3.32 x 10³¹ /1.24 x 10¹⁷
<u>t = 2.68 x 10¹⁴ years</u>
Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
<u>B) Have a shock wave</u>
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
