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Angelina_Jolie [31]
3 years ago
12

List how many significant figures the numbers contain. A. 205 cm. B. .00004 cm. C. 20 cm. D. 20. Cm.

Physics
1 answer:
stiv31 [10]3 years ago
7 0
A. 205 cm
   This has 3 significant figures: 2,0 and 5.

B. 0.00004 cm
   This has 1 significant figure: 4

C. 20 cm
   This has 2 significant figures: 2 and 0.

D. 20. cm
    This has 2 significant figures: 2 and 0 
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Answer:

a). Single replacement.

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What is the current in a 120V circuit if the resistance is 20Ω?
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We have: I=\frac{U}{R}=\frac{120}{20}=6A

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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

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3 years ago
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82ohms

Explanation:

The equivalent resistance in the circuit is 82ohms

Given parameters:

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Unknown:

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A resistor is an body in circuit that opposes the flow of electric current.

Resistors are usually connected in circuit and in series arrangement.

When resistors are connected in series, they have the same current passing through them.

Equivalent resistance is the sum of each of the connected resistors

Equivalent resistance = R1 + R2 = 50 + 32 = 82ohms

learn more:

Circuits brainly.com/question/2364338

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