Who is responsible for conducting a hazard assessment?

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago

technician a says that dirt bypassing the filter on many common rail injectors can cause an injector to stick open and continuou

NNADVOKAT [17]

Technician a is correct because he says that Many common rail injectors filters can be bypassed by dirt, which can lead to an injector sticking open and continuously fueling a cylinder.

Coalescence is used to separate the water and fuel. To the fuel injector cleaning kit, fasten your air compressor. Diesel engines run at compression ratios that are greater than those of gasoline engines. greater ratio compared to gasoline engines. increased thermal expansion as a result. more fuel energy that is transformed into usable power. The great benefit of using a dry cylinder sleeve is that by quickly installing new sleeves, the cylinder block can be quickly restored to its original specifications. Vacuum drying can be used to get rid of small amounts of water. A nozzle is used to spray the fuel into the vacuum chamber of engines. Air and unsolved free water are taken out of the oil. The fuel is evenly dispersed, which facilitates efficient drying.

Learn more about injectors here:

brainly.com/question/27969202

#SPJ4

3 0

11 months ago

Three groups of students are given study outlines for 6 weeks. One group studies 2 hours a night, a second group studies 1 hour

katrin2010 [14]

Answer:

The constant here is the study outline

Explanation:

In scientific research, the constant variable is that part/variable of the experiment that does not change or is set not to change. Examples include temperature, environment or height.

Assuming the scenery described in this question is an experiment. All the groups presented are bound by a constant during the experiment. The constant here is the study outline. The study outline provided to the students is not going to change.

NOTE: There could be confusion as regards the answer being the final exam grade but that will be the dependent variable as that will be the outcome of the experiment while the time spent to study will be the independent variable.

8 0

2 years ago

Search and destroy christian

Setler [38]

Answer: ummmm wut?

Explanation:

3 0

3 years ago

Read 2 more answers

A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,

juin [17]

Answer:

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

- Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE as follows:

stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B and A_ac: cross sectional area of member AE.

cos(Q) = 4 / 5 ..... From figure ( trigonometry )

A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

stress_ae = 2*(4/5) / 3*10^-4

stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

(M)_c = P*6 - F_eb*3

0 = P*6 - F_eb*3

F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

P - F_eb - F_c = 0

F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

A_pin = pi*d^2 / 4

A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

shear stress = 0.5*P / 2*A_pin

shear stress = 0.5*2 / 2*2.8353*10^-4

shear stress = 1.763 MPa

7 0

3 years ago