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a_sh-v [17]
3 years ago
15

When adjusting the valve clearances on the engine, Technician A says the valves are closed when the valve lifter or follower is

on the inner base circle of the camshaft lobe. Technician B says that valves are always adjusted when closed while the lifter or follower is on the camshaft lobe nose. Who is correct?
A. Technician A only
B. Technician B only
C. Both A and B
D. Neither A nor B
Physics
1 answer:
andre [41]3 years ago
7 0

Answer:

A. Technician A only

Explanation:

The camshaft opening ramp is what opens the valve, it starts at the basement of the inner circle and closes at the nose while the camshaft closing ramp allows for closing a valve starts at the nose and closes at the inner base circle. Valve clearance on an engine are performed when closing a valve, and is completed when the valve lifter or follower is on the inner base circle of the camshaft lobe.

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Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells
barxatty [35]

Answer:

a

The mass of blood is m= 5.7876kg

b

The number of blood cells is  N_t=1.04*10^{13}

Explanation:

From the question we are told that

         The volume of blood  is  V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3

         The density of the blood is  \rho_b = 1060 kg/m^3

         % of blood  that is  cell is  = 45.0%

        % of the blood that is  plasma is  = 55.0%

        density of blood cell is  \rho_d = 1125kg/m^3

        % of cell that are white is  = 1%

        % of cell that is red is  = 99%

        The diameter of the red blood cell is  = 7.5 \mu m = 7.5*10^{-6}m

         The radius of the red blood cell is  = \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m

Generally the mass is mathematically  represented as

               m = \rho_b * V_b

Substituting value

            m = 1060 * 0.00546

               m= 5.7876kg

Mass of cell is m_c = 45% of m

                         = 0.45 * 5,7876

                         = 2.60442 kg

The volume of cells is V_c = \frac{m_c}{\rho_d}

                                      = \frac{2.60442}{1125}

                                      = 2.315 *10^{-3} m^3

The volume of white blood cell is V_w = 1% of volume of cells

                                                         = \frac{1}{100} * 2.315*10^{-3}

                                                       = 2.315*10^{-5}m^3

The volume of a single cell is V_s = 4 \pi r^3

                                                                        = 4*(3.142) * (3.75*10^{-6})^3

                                                                        = 2.21*10^{-16}m^3

The volume of red blood cells is V_r = V_c - V_w

                                                           =2.315*10^{-3} - 2.315*10^{-5}

                                                           = 2.29*10^{-3}m^3

The number of red blood cell is  = \frac{V_r}{V_s}

                                                     = \frac{2.29 *10^{-3}}{2.21*10^{-16}}

                                                    = 1.037*10^{13}

The Number of white blood cell is   =\frac{V_w}{V_s}

                                                          = \frac{2.315 * 10^{-5}}{2.21*10^{-16}}

                                                          = 1.04*10^{11}

The total number of blood cells is  N_t= 1.037*10^{13} + 1.04*10^{11}

                                                        N_t=1.04*10^{13}

6 0
3 years ago
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Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

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Using this formula Vj = V; + at, If a vehicle starts from rest
Furkat [3]

Answer:

<em>The final speed of the vehicle is 36 m/s</em>

Explanation:

<u>Uniform Acceleration</u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+a.t

Where:

vf  = Final speed

vo = Initial speed

a   = Constant acceleration

t   = Elapsed time

The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:

v_f=0+4.5*8

v_f=36\ m/s

The final speed of the vehicle is 36 m/s

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