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Lynna [10]
4 years ago
15

An automobile traveling along a straight road increases its speed from 72 ft/s to 84 ft/s in 180 ft. if the acceleration is cons

tant, how much time elapses while the auto moves the 180 ft? answer in units of s.
Physics
1 answer:
Nikolay [14]4 years ago
5 0
The equation that would allow us to calculate for the acceleration given the distance is written below,

      a = (Vf² - Vo²) / 2d

where a is the acceleration, Vf is the final velocity, Vo is the initial velocity, and d is distance. 

Substituting the known values,
    a = ((84 ft/s)² - (72 ft/s)²) / 2(180 ft) = 5.2 ft/s²

Then, the equation that would relate the initial velocity, distance, acceleration and time is calculated through the equation,
      
     d = Vot + 0.5at²

Substituting the known values,
    180 = 72(t) + 0.5(5.2)(t²)

The value of t from the equation is 2.3 s

<em>ANSWER: 2.3 s</em>
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A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and th
tamaranim1 [39]

Answer:

<em>The power required by the pump is nearly 230.588 kW</em>

Explanation:

Flow rate of the pump Q = 1 m^3/s

the head flow H = 20 m

specific weight of water γ = 9800 N/m^3

efficiency of the pump η = 85%

First note that specific gravity of water is the product of the density of water and acceleration due to gravity.

γ = ρg

where ρ is density. For water its value is 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

The power to lift this water at this rate will be gotten from the equation

P = ρgQH

but ρg = γ

therefore,

P = γQH

imputing values, we'll have

P = 9800 x 1 x 20 = 196000 W

But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.

we can say that

196000 W is 85% of the power of the pump power needed, therefore

196000 = 85% of P_{p}

where P_{p} is the power of the pump needed

85% = 0.85

196000 = 0.85P_{p}

P_{p} = 196000/0.85 = 230588.24 W

<em>Pump power = 230.588 kW</em>

3 0
3 years ago
(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?
Mrrafil [7]

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization.  Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750  ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

5 0
4 years ago
3. Which part of the back seat of a car—the left side, the right side, or the
Nina [5.8K]

Answer:

The sides

Explanation:

Because there's a seat in front of the child to avoid him/her from flying

7 0
3 years ago
A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=\dfrac{1}{2}kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

U_{i}+U'_{i}=U_{f}+U'_{f}

\dfrac{1}{2}kx^2+0=0+mgh

h=\dfrac{kx^2}{2mg}

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}

h=11.653\ m

Hence, The maximum height above the point of release is 11.653 m.

3 0
3 years ago
As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a
Annette [7]
Im pretty sure it’s a because it makes more sense you know?.
5 0
3 years ago
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