To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

$P = \rho h$

Here,

h = Depth average

$\rho$ = Density

Moreover,

$\text{Density of water}= \rho = 62.4lb/ft^3$

Replacing,

$P = (62.4lb/ft^3)(\frac{35}{2}ft)$

$P = 1092 lb/ft^2$

Finally the force

$\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}$

$F = 1092lb/ft^2(101ft*52ft)$

$F = 5.735*10^6lbf$

6 0

2 years ago

Stopping distance of vehicles When brakes are applied to a moving vehicle, the distance it travels before stopping is called sto

DochEvi [55]

Answer:

Stopping distance = 40m

Explanation:

Given the following :

Initial speed of vehicle before applying brakes = 72km/hr

Converting km/hr to m/s:

72km/hr = [(72 * 1000)m] / (60 * 60)

72km/hr = 72,000m / 3600s

72km/hr = 20m/s

Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2

From the 3rd equation of motion:

v^2 = u^2 + 2as

Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance

Final velocity when the car stops will be 0

Therefore ;

v^2 = u^2 + 2as

0 = 20^2 + 2(-5)(s)

0 = 400 - 10s

10s = 400

s = 400/10

s = 40m

Therefore, the stopping distance of the car = 40 meters

7 0

2 years ago

Question 1: Bicycle pump

belka [17]

Answer: The pressure of the air when compressed is $5.0\times 10^5Pa$

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = $1.0\times 10^5Pa$

$P_2$ = final pressure of gas =?

$V_1$ = initial volume of gas = V

$V_2$ = final volume of gas = $\frac{V}{5}$

$1.0\times 10^5\times V=P_2\times \frac{V}{5}$

$P_2=5.0\times 10^5Pa$

Therefore, the pressure of the air when compressed is $5.0\times 10^5Pa$

4 0

3 years ago