A ball is dropped from a window sill and hits the ground in 4.1s. What is the final velocity?

A woman is sitting at a bus stop when an ambulance with a siren wailing at 317 Hz approaches at 69 miles per hour (mph). Assume

xxMikexx [17]

Answer:

a) 30.84m/s

b) 348.32Hz

c) 32.34m/s

d) 289.69Hz

Explanation:

a) If 1 mile=1609,34m, and 1 hour=3600 seconds, then 69mph=69*1609.34m/3600s=30.84m/s

b) Based on Doppler effect:

/*I will take as positive direction the vector $\vec r_{observer}-\vec r_{emiter}$ */

$f_{observed}=(\frac{v_{sound}-v_{observed}}{v_{sound}-v_{emited}})f_{emited}$

$f_{observed}=(\frac{343m/s-0m/s}{343m/s-30.84m/s})317Hz=348.32Hz$

c) $350Hz=(\frac{343m/s-0m/s}{343m/s-v_{ambulance}})317Hz, V_{ambulance}=343m/s-\frac{317Hz}{350Hz}.343m/s=32.34m/s$

d) $f_{observed}=(\frac{343m/s-0m/s}{343m/s+32.34m/s})317Hz=289.69Hz$

3 0

4 years ago

4 points

zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

$I = k\frac{1}{r^2}$

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

$I_2 = k\frac{1}{(2r)^2}$

Dividing both the equation we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4} = 25lx$

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

$I_2 = k\frac{1}{(3r)^2}$

Dividing equation 1 and 3 we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9} = 11.11lx$

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0

3 years ago

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If

frez [133]

Answer:

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2

So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2

from continuity we have ρ×A1×v1 = ρ×A2×v2

v2 = v1×A1/A2

and

r2 = 2×r1

then:

A2 = 4×A1

so,

v2 = (v1)/4

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)

= 9.75×10^4 Pa

= 9.8×10^4 Pa

Therefore, the pressure in the wider section is 9.8×10^4 Pa

5 0

3 years ago

Suppose you have 20% of 20/100 or 1/5. Describe how you can solve the problem using 1/5 instead of 20%.

grin007 [14]

Answer:

To calculate Percent in fraction You need to do following steps.

Divide number by 100.

=20/100

=1/5

6 0

3 years ago

Read 2 more answers

¡¡first convert revolutions per second to

katovenus [111]

Answer:

a. 5.23 m/s² b. 44.23 N

Explanation:

a. What is the centripetal acceleration of the hammer?

The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.

Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.

So a = rω²

= 1.4 m × (3.74 rad/s)²

= 5.23 m/s²

b. What is the tension in the chain?

The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.

So, T = 8.45 kg × 5.23 m/s²

= 44.23 N

6 0

3 years ago