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3 years ago

Potential difference must be measured in parallel within a circuit, using a device called a voltmeter

Physics

1 answer:

grin007 [14]3 years ago

4 0

Answer:

This is true

Explanation:

This is true because in other To measure the potential difference that is across a circuit, we have to place the device called the voltmeter to be in parallel withe the circuit, so as to get a measurement of the difference in energy from one side of the circuit to the other side. We can also refer to the Potential difference as the voltage and it is measured in volts using a voltmeter

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What happens to the electronegativity of elements as you move from bottom left to upper right across the periodic table?

Paraphin [41]

It increases. As it moves it <span>increases while the movement is in process.

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3 0

3 years ago

A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car

asambeis [7]

Answer:

350 miles

Explanation:

When the car starts 2 hours later, the train would have a head start of

50 * 2 = 100 miles

The speed of the car relative to the train is

70 - 50 = 20 mi/hr

For the car to catch up with the train, it must cover the 100 miles difference at the rate of 20mi/hr. So the time it would need to cover this difference is

100 / 20 = 5 hours

After 5 hours, the car would have traveled a distance of

5 * 70 = 350 miles which is also the distance from the station to where the car catches up

6 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

A student performing a double-slit experiment is using a green laser with a wavelength of 550 nm. She is confused when the m = 5

sweet [91]

Answer:

d = 52 μm

Explanation:

given,

wavelength of the light source (λ)= 550 nm

distance to form interference pattern(D) = 1.5 m

y = 1.6 cm = 0.016 m

width of the slits = ?

now, using displacement formula

$y = \dfrac{m\lambda\ D}{d}$

for the first maxima, m = 1

$d = \dfrac{1\times \lambda\ D}{y}$

$d = \dfrac{1\times 550 \times 10^{-9}\times 1.5}{0.016}$

d = 5.2 x 10⁻⁶ m

d = 52 μm

hence, the width of her slits is equal to d = 52 μm

3 0

3 years ago

the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?

Inessa [10]

Via the half-life equation:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where the time elapse is 11,460 year and the half-life is 5,730 years.

$A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}$

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

6 0

3 years ago