Answer:
When a horse pull a cart the action is on?
A horse is harnessed to a cart. If the horse tries to pull the cart, the horse must exert a force on the cart. By Newton's third law the cart must then exert an equal and opposite force on the horse. Newton's second law tells us that acceleration is equal to the net force divided by the mass of the system.
Explanation:
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑
= mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
A. I believe, lmk if I’m right
Answer: Attached below is the missing diagram
answer :
A) 1) Wr > WI, 2) Qc' > Qc
B) 1) QH' > QH, 2) Qc' > Qc
Explanation:
л = w / QH = 1 - Qc / QH and QH = w + Qc
<u>A) each cycle receives same amount of energy by heat transfer</u>
<u>(</u> Given that ; Л1 = 1/3 ЛR )
<em>1) develops greater bet work </em>
WR develops greater work ( i.e. Wr > WI )
<em>2) discharges greater energy by heat transfer</em>
Qc' > Qc
solution attached below
<u>B) If Each cycle develops the same net work </u>
<em>1) Receives greater net energy by heat transfer from hot reservoir</em>
QH' > QH ( solution is attached below )
<em>2) discharges greater energy by heat transfer to the cold reservoir</em>
Qc' > Qc
solution attached below