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Zolol [24]
3 years ago
10

18. A 2.0-ohm resistor is connected in a series with a 20.0 -V battery and a three-branch parallel network with branches whose r

esistance are 8.0 ohms each. Ignoring the battery’s internal resistance, what is the current in the batter? Show your work.

Physics
2 answers:
jonny [76]3 years ago
5 0
Here is the answer.  First get the impedance of the 3 parallel 8 ohm resistors, then add in the series resistor and solve with Ohm's Law

Lapatulllka [165]3 years ago
5 0

the correct answer would be 4.28

8ll8ll8 =1/ 1/8 /1/8 /1/8 = 2.67

2+2.67=4.67

20v/4.67=4.28A

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I need help with life a child is adopted and I’m a duck where is my left shoe
Pavel [41]

just swim In water and find your shoe

8 0
3 years ago
Read 2 more answers
After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average
lara [203]

Answer:

compression of spring is x = 0.12 m

Assumed k = 160,000 N/m ........ Truck's suspension system

Explanation:

Given:

- The mass of average person m_p = 69 kg

- Total number of persons n_p = 27

- The mass of each goat m_g = 15 kg

- The total number of goats n_g = 3

- The mass of each chicken m_c = 3 kg

- The total number of goats n_c = 5

- The total mass of bananas m_b = 25 kg

Find:

How much are the springs compressed?

Solution:

- Using equilibrium equation on the taptap in vertical direction:

                                 F_net = F_spring - F_weight = 0

- Compute the force due to all the weights on the taptap:

                                F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b)*9.81

                                F_weight = (69*27 + 3*15 + 5*3 + 25)*9.81  

                                F_weight = 19109.88 N

- The restoring force of a spring is given by:

                                F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

                                 F_weight = F_spring

                                 19109.88 = k*x

                                 x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

                                 x = 19109.88 / 160,000

                                 x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.

8 0
3 years ago
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0
3 years ago
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

4 0
3 years ago
Read 2 more answers
During 57 seconds of use, 330 C of charge flow through a microwave oven. Compute the size of the electric current.
swat32

Answer:

5.78amps

Explanation:

Given data

Time t= 57 seconds

Charge Q= 330C

Current I= ??

The expression for the electric current is given as

Q= It

Substituting we have

330= I*57

I= 330/57

I=5.78 amps

Hence the current is 5.78amps

3 0
3 years ago
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