A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

7 0

3 years ago

A fishing boat accidentally spills 3.0 barrels of diesel oil into the ocean. each barrel contains 42 gallons. if the oil film on

9966 [12]

Number of barrels are 3.0. Each barrel contains 42 gallons of oil. Thus, total volume of oil will be 42×3=126 gallons.

Converting gallons into m^{3}

1 gallon=0.00378 m^{3}

Thus, 126 gallons=0.4769 m^{3}

Thickness of oil film is 2.5\times 10^{2} nm, converting it into meters as follows:

1 nm=10^{-9} m

Thus,

2.5\times 10^{2} nm=1.5\times 10^{-7}m

Now, volume V of oil is related to area A and thickness T as follows:

V=A×T

rearranging,

A=\frac{V}{T}=\frac{0.4769 m^{3}}{2\times 10^{-7}m}=2.38\times 10^{6}m^{2}

Thus, square meters of oil will be 2.38\times 10^{6}m^{2}

7 0

3 years ago

A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat

RSB [31]

Answer:

Explanation:

The question states that the chairs are spaced 20 m apart through a length of 1 km, or say, 1000 m.

It also does say that each chair weighs 250 kg, and as such the load is

M = 50 * 250

M = 12500.

Taking into consideration, the initial and final heights, we have

h1 = 0, h2 = 200 m

The work needed to raise the chairs,

W = mgh, where h = h2 - h1

W = 12500 * 9.81 * (200 - 0)

W = 2.54*10^7 J

The work is done at a rate of 10 km/h, and at a distance of 1 km, time taken would be

t = 1/10 = 0.1 h or say, 360 s

The power needed thus, is

P = W/t

P = 2.54*10^7 / 360

P = 68125 W, or 68 kW

Initial velocity, u = 0 m/s

Final velocity, v = 10 km/h = 2.78 m/s

Startup time, t is 17 s

Acceleration during the startup then is

a = (v - u)/t

a = 2.78/17

a = 0.163 m/s²

The power needed for the acceleration is

P = ½m [(v² - u²)/t]

P = ½ * 12500 * [2.78²/17]

P = 6250 * 0.455

P = 2844 W

3 0

3 years ago

The fact that electric charges return to the source of the current is an example of:

Marizza181 [45]

The law of conservation of charge.

6 0

2 years ago

The energy required to dislodge electrons from cesium metal via the photoelectric effect is 207 kJ/mol . What wavelength (in nm)

ki77a [65]

Answer:

So wavelength in nm will be $\lambda =0.956\times 10^{-19}nm$

Explanation:

We have given that the energy = 207 KJ/mol $=207\times 10^3j/mol$

Speed of light $c=3\times 10^8m/sec$

Plank's constant $h=6.6\times 10^{-34}J-s$

According to plank's rule energy of the photon is given by

$E=\frac{hc}{\lambda }$

$207\times 10^3=\frac{6.6\times 10^{-34}\times 3\times 10^8}{\lambda }$

$\lambda =0.956\times 10^{-28}m$

So wavelength in nm will be $\lambda =0.956\times 10^{-19}nm$

6 0

3 years ago