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Neko [114]
3 years ago
7

Based on the data table,which car won the race?

Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

car one

Explanation:

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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
3 years ago
A climber using bottled oxygen accidentally drops the oxygen bottle from an altitude of 4500 m. If the bottle fell straight down
Trava [24]

Answer:

v= 300 m/s

Explanation:

Given that

altitude ,h= 4500 m

The mass ,m = 3 kg

Lets take acceleration due to gravity , g= 10 m/s²

The speed before impact at sea  level =  v

Initial speed ,u = 0 m/s

We know that

v²=u²+2 g h

v=final speed

u=initial speed

h=height

Now by putting the values in the above equation

v² = 0²+ 2 x 10 x 4500

v²=90000

v= 300 m/s

Therefore the speed at sea level will be 300 m/s.

3 0
3 years ago
An object carries a +15.5 uC charge.
abruzzese [7]

Answer:

3.67 N

Explanation:

From the question given above, the following data were obtained:

Charge of 1st object (q₁) = +15.5 μC

Charge of 2nd object (q₂) = –7.25 μC

Distance apart (r) = 0.525 m

Force (F) =?

Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:

For the 1st object

1 μC = 1×10¯⁶ C

Therefore,

15.5 μC = 15.5 × 1×10¯⁶

15.5 μC = 15.5×10¯⁶ C

For the 2nd object:

1 μC = 1×10¯⁶ C

Therefore,

–7.25 μC = –7.25 × 1×10¯⁶

–7.25 μC = –7.25×10¯⁶ C

Finally, we shall determine the force. This can be obtained as follow:

Charge of 1st object (q₁) = +15.5×10¯⁶ C

Charge of 2nd object (q₂) = –7.25×10¯⁶ C

Distance apart (r) = 0.525 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²

F = 3.67 N

Therefore, the force on the object is 3.67 N

4 0
3 years ago
HELP PLEASE PHYSICS !!!
astraxan [27]
C is the answer I think good luck
5 0
3 years ago
A CO molecule has an intrinsic dipole moment whose magnitude is 4 X 10^-31 C*m. If the separation between the atoms is 0.11 nm,
aleksandrvk [35]
4x10^-31/(0.11x10^-9)= 3,63 x 10^-21
5 0
4 years ago
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