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baherus [9]
2 years ago
6

On a Vernier Caliper, how do you know which mark to use on the very top scale?

Physics
1 answer:
madreJ [45]2 years ago
4 0

<u>Answer</u>

To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use<em>. </em>


<u>Explanation</u>

A vernier caliper is an instrument that is used to measure the diameter of small circular objects such as diameter of a wires, thickness of an iron sheet.

The objects to be measured is place between the jaws of the calipers.

The vernier scale has two scales, the vernier scale and the main scale which is the very top scale.<em> To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use. </em>

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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0
3 years ago
Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?
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The brand of car wax
4 0
2 years ago
If a 4.0Ω resistor, a 6.0Ω resistor, and an 8.0Ω resistor are connected in parallel across a 12 volt battery, what is the total
allsm [11]

Answer:

Explanation:

6.7 amps

7 0
2 years ago
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
3 years ago
A train travelled 500 meters in 25 seconds
KIM [24]

\huge\mathfrak\red {Answer:}

<h2>Speed = Distance/Time</h2>

If a train travelled 500 meters in 25 seconds then,

Speed = 500m/25sec

<h2>→ 20 m/sec</h2>

\mathfrak\purple {Hope\: this\: helps\: you...}

7 0
2 years ago
Read 2 more answers
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