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3 years ago

10

A student walks the hallway for 25 m stops to talk and continues down the hallway another 10 m what is the distance and displace

ment

1 answer:

brilliants [131]3 years ago

3 0

Answer:

Distance = displacement = 35m

Explanation:

The distance of the student is how far he has gone.

Distance = 25m + 10m

Distance = 35m

Displacement is the distance specified in specific direction. Since the student walk in the sane direction, thence the displacement is also 35m

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If 710- nm and 660- nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wa

alexira [117]

0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.

<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.5m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3276.92 x 10⁻⁶ m

= 3.276x 10⁻³ m

= 3.276mm .

For λ = 660 nm

position = 2 λ D / d

λ = 660 nm , D = 1.5 m

d = .65 x 10⁻³

position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3046.15 x 10⁻⁶ m

= 3.046 x 10⁻³ m

= 3.046 mm .

Difference between their position

= 3.276mm ₋ 3.046 mm

= 0.23 mm .

To know more about Fringes refer to: brainly.com/question/15649748

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7 0

2 years ago

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

3 years ago

In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug

Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

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3 years ago

The cytoplasm of a cell consists mainly of water.

ankoles [38]

Deal! The answer is true

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3 years ago

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An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh

lara [203]

Answer:

$a_y = 4.9\ m/s^2$

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

$\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}$

$\dfrac{dz}{dy}=-0.5$

The differential form with the acceleration

$\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}$

$-0.5=-\dfrac{a_y}{a_z + g}$

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

$0.5=\dfrac{a_y}{0 + 9.8}$

$a_y = 9.8\times 0.5$

$a_y = 4.9\ m/s^2$

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

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3 years ago