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Ksju [112]
4 years ago
14

Technician A says that outer tie rod ends should be replaced in pairs, even if only one is worn. Technician B says that inner ti

e rod ends should be replaced in pairs, even if only one is worn. Which technician is correct?
A. Technicain A only
B. Technician B only
C. Both A and B
D. Neither A nor B
Physics
1 answer:
DENIUS [597]4 years ago
4 0

Answer:

C)Both A and B

Explanation:

It is best practice to replace the worn out tie rod with Pairs instead of a single tie rod.  It is not absolutely necessary, but many technicians insist on replacing both even if only one is worn out. This is because both have more or less the same load and sooner or later the other one is going to fail too. That's why technician A said that outer tie rod ends should be replaced in pairs, even if only one is worn. Technician B said that inner tie rod ends should be replaced in pairs, even if only one is worn. Both the technicians are correct

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A satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m. What is the acceleration of the satellite?
Anna11 [10]

A satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m. then the acceleration of the satellite would be 5.832 m/s²

<h3 /><h3>What is a uniform circular motion?</h3>

It is defined as motion when the object is moving in a circle with a constant speed and its velocity is changing with every moment because of the change of direction but the speed of the object is constant in a uniform circular motion

The acceleration during the uniform circular motion is given by the formula

a = v²/r

where a is the acceleration

v is the velocity of the object

r is the radius of the orbit

As given in the problem a satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m.

velocity of the satellite

v = 2πr/t

v= (2×3.14×2008177)/3685

v = 3422.34 m/s

acceleration of the satellite

a = v²/r

a =  3422.34²/2008177

a = 5.832 m/s²

Thus, the acceleration of the satellite would be 5.832 m/s²

Learn more about uniform circular motion from here

brainly.com/question/2285236

#SPJ1

7 0
2 years ago
A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling
sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

P_{1} = P_{2}

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

v_{1_{i}}: is the initial velocity of the ball 1 = 6.20 m/s

v_{2_{i}}: is the initial velocity of the ball 2 = 0 (it is at rest)

v_{1_{f}}: is the final velocity of the ball 1 =?

v_{2_{f}}: is the initial velocity of the ball 2 =?

m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}} (1)        

Now, by conservation of kinetic energy (since they collide elastically):

\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}          

m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}  (2)

By entering equation (1) into (2) we have:

m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}    

0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}            

By solving the above equation for v_{2_{f}}:

v_{2_{f}} = 3.1 m/s

Now, v_{1_{f}} can be calculated with equation (1):

v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s

The minus sign of v_{1_{f}} means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!                  

5 0
3 years ago
Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30
vekshin1

66.7 Watts

Explanation:

Let R_{1}=1.0 ohms, R_{2}=3.0ohms and R_{3}=6.0\:ohms. Since R_{2} and R_{3} are in parallel, their combined resistance R_{23} is given by

\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}

or

R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms

The total current flowing through the circuit <em>I</em> is given

I=\dfrac{V_{s}}{R_{Total}}

where

R_{Total}=R_{1}+R_{23}= 3.0\:ohms

Therefore, the total current through the circuit is

I=\dfrac{30\:V}{3.0\:ohms}=10\:A

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor V_{1}:

V_{1}=(10\:A)(1.0\:ohms)=10\:V

This means that voltage drop across the 6-ohm resistor V_{3} is 20 V. The power dissipated <em>P</em> by the 6-ohm resitor is given by

P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W

4 0
3 years ago
Please help me I am desperate this is due now
elena55 [62]

Answer:

Distance travelled is 7 meters and the displacement is 3 meters

7 0
3 years ago
Read 2 more answers
A 75 kg skydiver can be modeled as a rectangular "box" with dimensions 20 cm * 40 cm * 180 cm. what is his terminal speed if he
Zina [86]

Terminal speed=136.7 m/s

the formula for the terminal speed is given by :

V= \sqrt{\frac{2 mg}{\rho a C}}

m= mass=75 kg

a= area=0.2 x 0.4=0.08 m²

ρ= density of air=1.23 kg/m³

C= drag constant =0.8

so V= \sqrt{\frac{2 *75*9.8}{1.23*0.08*0.8}}

v= 136.7 m/s

4 0
3 years ago
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