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faltersainse [42]
3 years ago
8

The force, F, of the wind blowing against a building is given by where V is the wind speed, rho the density of the air, A the cr

oss-sectional area of the building, and CD is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.
Physics
1 answer:
Sveta_85 [38]3 years ago
7 0

Answer:

dimensions of the drag coefficient is [M^0 L^0 T^0]

Drag coefficient is a dimensionless quantity

Explanation:

force is given byF=\frac{C_{D}  \rho V^2 A}{2}

we get expression for drag coefficient C_{D} =\frac{2F}{\rho V^2 A}

By substituting the dimensions  of the F,V,A and density , we get

C_{D} =\frac{[F]}{[\rho ][V]^2[A]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L T^{-1}]^2[L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L^2 T^{-2}][L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[MLT^{-2}]}\\C_{D}=[M^0 L^0 T^0]

Drag coefficient is a dimensionless

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Two slits are illuminated with green light (λ = 540 nm). The slits are 0.05 mm apart and the distance to the screen is 1.5 m. At
zhenek [66]

Answer:

0.21486 mm

Explanation:

The formula for the maximum intensity is given by;

I = I_o•cos²(Φ/2)

Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)

Where;

y is the distance from the central maximum

d is the distance between the slits

λ is the wavelength

L is the distance to the screen

Thus;

I = I_o•πdy/(λL)

We are given;

d = 0.05 mm = 0.5 × 10^(-3) m

λ = 540 nm = 540 × 10^(-9) m

L = 1.25 m

I/I_o = 50% = 0.5

From earlier, we saw that;

I = I_o•πdy/(λL)

We have I/I_o = 0.5

Thus;

I/I_o = πdy/(λL)

Plugging in the relevant values;

0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)

Making y the subject, we have;

y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))

y = 0.00021486 m

Converting to mm, we have;

y = 0.21486 mm

7 0
3 years ago
Objects fall near the surface of the earth with a constant downward acceleration of 10 m/s2 . Suppose a falling object is moving
Papessa [141]

Answer:

The final velocity of the object after 2 seconds is 30 m/s

Explanation:

Given;

constant downward acceleration, a =  10 m/s²

initial velocity of the object falling down, v = 10 m/s

time of fall, t = 2 s

The final velocity of the object is given by;

v = u + at

where;

v is the final velocity

v = 10 + (10)(2)

v = 10 + 20

v = 30 m/s

Therefore, the final velocity of the object after 2 seconds is 30 m/s

3 0
3 years ago
A 3.0 kg object is loaded into a toy spring gun, and the spring has a spring constant 750N/m. The object compresses the spring b
kiruha [24]

Answer:

Explanation:

mass of object, m = 3 kg

spring constant, K = 750 n/m

compression, x = 8 cm = 0.08 m

angle of gun, θ = 30°

(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.

(b) Let v be th evelocity of ball at the tim eof launch.

by using the conservation of energy

1/2 Kx² = 1/2 mv²

750 x 0.08 x 0.08 = 3 x v²

v = 1.265 m/s

By use of the formula of maximum height

h = \frac{v^{2}Sin^{2}\theta}{2g}

h = \frac{1.265^{2}Sin^{2}30}{2\times 9.8}

h = 0.02 m

h = 2 cm  

4 0
3 years ago
A thin plastic rod of length 2.5 m is rubbed all over with wool, and acquires a charge of 75 nC, distributed uniformly over its
Colt1911 [192]

Answer:

find out the le

Explanation:

3 0
3 years ago
Which are ways to improve the design of this experiment? Check all that apply.
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4 years ago
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