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2 years ago

What are the advantages of parallel circuits? Check all that apply

Physics

2 answers:

Bezzdna [24]2 years ago

7 0

Answer:

If one bulb goes out, other bulbs stays lit.

If there is break in one branch of the circuit, current can still flow through other branches.

Parallel circuits are simple to design and build.

Explanation:

In parallel circuits all the components and resistors are connected to common terminal or common supply.

So here if we disconnect one of the branch of the circuit then it will not affect the other branches of the circuit so it will not affect the current of other branches.

So here it is very easy to build a parallel circuit and if one of the branch of the circuit goes out then it will not affect the other branches.

aleksandrvk [35]2 years ago

6 0

The advantages of parrallel circuits are

If one bulb goes out , the other bulb stays lit

If there is a break in one branch of the circuit, current can still flow through the other branches of the circuit

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3 years ago

PHYSICS

quester [9]

Answer:

c) 8.3 x 10^24 molecules

Explanation:

First of all, we convert the volume of the glass of water from mL to $cm^3$ :

$V=250 mL = 250 cm^3$

The relationship between mass and volume is:

$d=\frac{M}{V}$

where

$d=1.00 g/cm^3$ is the density of water

M is the total mass of the glass of water

$V=250 cm^3$ is the volume of the water in the glass

Solving for M,

$M=dV=(1.00)(250)=250 g = 0.250 kg$

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$M=Nm$

where

N is the number of molecules in the glass of water

Solving for N, we find:

$N=\frac{M}{m}=\frac{0.250}{3.0\cdot 10^{-26}}=8.3\cdot 10^{24}$

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3 years ago

Children are told to avoid standing too close to a rapidly moving train because they might get sucked under it. Is this possible

storchak [24]

Answer:

no its not like the undertow in the ocean

Explanation:

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3 years ago

75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is

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Answer:

Explanation:

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3 years ago

Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

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3 years ago