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2 years ago

A gannet is a seabird that fishes by diving from a great height

1 answer:

3 0

For constant acceleration along a given direction, we can relate acceleration, velocity and position with the following equation that doesn't involve time:
$v^{2}= v_{0}^{2}+2a(x- x_{0})$
In this equation x is the final position, which we take to be 0. Also the initial velocity Vo is zero. Thus the equation simplifies to
$v^{2}= 2a(- x_{0})$
Putting in v=32m/s, a=-9.81m/s^2 gives
$32^{2}= 2(-9.81)(- x_{0})
$
$\frac{32^{2}}{2*9.81}= x_{0}=52.19m$

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Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi

Serga [27]

Answer:

Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)

Explanation:

Given

Hf = head loss
f = friction factor
L = Length of the pipe = 360 m
V = Flow velocity, m/s
D = Pipe diameter = 0.12 m
g = Gravitational acceleration, m/s^2
Re = Reynolds's Number
rho = Density =998 kg/m^3
μ = Viscosity = 0.001 kg/m-s
Z = Elevation Difference = 60 m

Calculations

Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

7 0

2 years ago

An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars

Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

$W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2$

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

$W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}$

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

$P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}$

the car did not exceed the speed limit

Hope this helps!!

7 0

2 years ago

Read 2 more answers

The normal boiling point of cyclohexane is 81.0 oC. What is the vapor pressure of cyclohexane at 81.0 oC?

Aloiza [94]

Answer:

The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

Explanation:

Given that,

Boiling point = 81.0°C

Atmospheric pressure :

Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.

The value of atmospheric pressure is

$P=101325\ Pa$

Vapor pressure :

Vapor pressure is equal to the atmospheric pressure.

Hence, The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

6 0

2 years ago

Torque, a car manufacturer, aims to make its slogan "Burn Rubber" known to every individual in a way that they immediately assoc

mylen [45]

Answer:

A customer is about to buy a limited edition sports car from Torque. It is most likely that the customer will have d. limited problem solving, in which consumers decision rules to purchase are simple, and are not motivated to search for information about other optons, deciding to buy the car.

Explanation:

a. advertising clutter is the big amount of ad-messages that the consumer is exposed to everyday.

b. high involvement product or purchases features many variables to be considered by the customer before getting to a decision.

c. cognitive dissonance or conflicting attitudes causing mental discomfort.

7 0

2 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

2 years ago