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3 years ago

If the voltage impressed across a circuit is held constant while the resistance is halved, what change occur?

Physics

1 answer:

egoroff_w [7]3 years ago

3 0

Answer:

The current doubles

Explanation:

In a circuit, Ohm's law gives the relationship between voltage, current and resistance:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

In this problem,

V is held constant

R is halved: $R'=\frac{R}{2}$

Therefore, the new current is

$I'=\frac{V}{R'}=\frac{V}{R/2}=2\frac{V}{R}=2I$

So, the current doubles.

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A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system applicat

umka21 [38]

Answer:

$Av_2 =24\ V/V$

Explanation:

given,

op-amp circuit with a gain of = (Av₁) = 96 V/V

Band width = (Bw₁) = 8 kHz

Required bandwidth(Bw₂) = 32 kHz

Highest gain available =(Av₂) = ?

For the given system Bandwidth product is constant

Av₁ Bw₁ = Av₂ Bw₂

96 x 8 = Av₂ x 32

$Av_2= \dfrac{96\times 8}{32}$

$Av_2 =24\ V/V$

the highest gain available under these conditions $Av_2 =24\ V/V$

6 0

3 years ago

A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli

Gre4nikov [31]

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble $m_1=41gram=0.041kg$

Velocity of marble $v_1=2.30m/sec$

Its collides with other marble of mass 25 gram

So mass of other marble $m_2=25gram=0.025kg$

Second marble is at so $v_2=0m/sec$

We have to find the velocity of second marble

From momentum conservation we know that

$m_1v_1+m_2v_2=(m_!+m_2)v$ , here v is common velocity of both marble after collision

So $0.041\times 2.30+0.025\times 0=(0.041+0.025)v$

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

6 0

3 years ago

Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea

notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

$v=u+at$

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

$0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s$

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

3 0

3 years ago

Read 2 more answers

Shanai drove 40 miles to the west, then turned around and drove 70 miles to the east. Draw vectors that show each segment of her

Dmitry [639]

Well she drove 30 more miles to the east than the west but I don’t understand what u are asking

4 0

2 years ago

A horizontal aquifer is overlain by 73.5 ft of water-saturated clay with a porosity of 0.4. The solid density of the clay partic

mr_godi [17]

Answer:

Explanation:

solution is in the attachment below

5 0

3 years ago