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puteri [66]
2 years ago
10

Jasmine makes a note card about series circuits. she records the points shown. a 2 column table with 1 row. the column on the le

ft is labeled advantages with entries (1) they are simple to design and build. the second column is labeled disadvantages with entries (2) each source of current used adds more current to the circuit, (3) the more bulbs are added the dimmer the bulbs are, (4) if one bulb goes out they all go out. which change will correct the error on her note card? switching the headings on the columns adding the title parallel circuits moving point (2) into the advantages column moving point (4) into the advantages column
Physics
1 answer:
Bess [88]2 years ago
7 0

Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

<h3>What is a parallel circuit?</h3>

A parallel circuit is a circuit in which the components are connected to a common junction. This implies that if one bulb goes out in a parallel connection, all the bulbs will go out.

As such, Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

Learn more about parallel connection: brainly.com/question/12400458

#SPJ4

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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the
zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

V = \frac{\pi d^2}{4}h

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

V_0 = \frac{\pi d^2}{4}H

V_f = \frac{\pi d^2}{4}h

We know as well by definition that

1gal = 3.785*10^{-3}m^3

Then we have for the statement that

V_f = V_0 -1gal

V_f = V_0 - 3.785*10^{-3}

Replacing the previous data

\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}

\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}

Solving to get h,

h = 1.99963m

Therefore the change is

\Delta h = H-h

\Delta h = 2- 1.99963

\Delta h = 3.7*10^{-4}m=0.37mm

Therefore te change in the height of the water in the tank is 0.37mm

4 0
3 years ago
What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?
sveta [45]
Momentum = (mv). 
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>
8 0
3 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0
3 years ago
Help with #3 please
Marizza181 [45]

Answer:

The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²

Explanation:

Data

Ta = 0.125 s

Tb = 0.08 s

Δtab = 0.5 s

distance = 1 cm

Process

1.- Calculate va

va = 1/0.125 = 8 cm/s

vb = 1/0.08 = 12.5 cm/s

2.- Calculate Δv

Δv = 12.5 - 8

Δv = 4.5 cm/s

3.- Calculate acceleration

a = Δv / Δt

a = 4.5/0.5

a = 9 cm/s²

4 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

7 0
3 years ago
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