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3 years ago

What symbols are used to explain displacement in physics

Physics

1 answer:

lidiya [134]3 years ago

4 0

Answer:

In this text the upper case Greek letter Δ (delta) always means “change in” whatever quantity follows it; thus, Δx means change in position. Always solve for displacement by subtracting initial position x0 from final position xf

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The scanner records the time from when a ultrasound wave is emitted to when its reflection is received

telo118 [61]

Explanation:

The scanner records the time from when a ultrasound wave is emitted to when its reflection is received. A technician calculates the depth of the reflection using the equation as :

$\text{depth}=\dfrac{1}{2}\times \text{speed of ultrasound in patient}\times \text{time recorded by scanner}$

The distance covered by ultrasonic wave when it was emitted and gets reflected is 2d. Speed is given by :

$v=\dfrac{d}{t}$

d is distance and t is time

Here, d = 2d

So, the factor of (1/2) is used because the distance covered by the wave is 2 times when it was emitted and received.

3 0

3 years ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

3 years ago

Please help me : (

Alex777 [14]

The correct answer is B) One atom of calcium and two atoms of chlorine. I just had the same question on my 9th grade science test today. I hope I didn't give the answer to you too late and I hope it helps you out

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4 years ago

Coffee or certain cola drinks can be used to test the effect of __________ on reaction time. What drug completes the sentence?

natulia [17]

Answer:

"caffeine" both products contain this compound

7 0

2 years ago

A light source of wavelength \lambdaλ illuminates a metal with a work function of \text{BE} = 2.00 ~\text{eV}BE=2.00 eV and ejec

dsp73

Answer:

1 eV

Explanation:

Given:

Work function, ∅ = 2.00 eV

Kinetic energy of the ejected of the electron, K.E = 4.0 eV

Now,

using the photoelectric equation , we have

Energy of the photon (E) = ∅ + K.E

also,

E = hc/λ

where, h is plank's constant

c is the speed of the light

λ is the wavelength

thus, we have

hc/λ = 2 + 4 = 6 eV

Energy of photon = 6eV

Now,

for the second case

λ' = 2λ

when Wavelength is doubled , E is halved

thus,

E' = hc/λ'

E' = hc/2λ

E' = E/2 = 6/2 = 3 eV

also,

E' = ∅ + KE '

thus on substituting the values,

3 = 2 + KE'

KE' = 1 eV

Hence, the maximum kinetic energy for the second case is 1 eV

5 0

3 years ago

What symbols are used to explain displacement in physics​

What symbols are used to explain displacement in physics