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iren2701 [21]
2 years ago
15

If she runs 3,223 meters each day and keeps this up for two weeks, how many miles will she have run

Physics
1 answer:
seropon [69]2 years ago
5 0

Answer:

28 miles

Explanation:

3,223 meters/day x 14 days = 45,122 meters

1 meter = 3.28 feet

45,122 meters = 148,000 feet

5280 feet = 1 mile

148,000 feet = 28 miles

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2 years ago
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A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t
vladimir2022 [97]

Answer

Radius of the wheel r = 2.1 m

Moment of inertia I = 2500 Kg m²

Tangential force applied F = 18 N

Time interval t = 16 s

Initial angular speed ω1 = 0

Final angular speed ω2 = ?

Let α be the angular acceleration.

Torque applied τ = Iα

                         F r = Iα

Angular acceleration α = F r/I

                                    = \dfrac{18\times 2.1}{2500}

                                    = 0.015 rad/s²

(a)From rotational kinematic relation

            Final angular speed ω₂ = ω₁ + αt

                                                 = 0 + (0.015 rad/s^2 * 16 s)

                                                 = 0.24 rad/s

(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²

                      = 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0

                      =  72 J

(c) Average power supplied by the child P = W/t = \dfrac{72}{16}

                                                                        = 4.5 watt        

8 0
3 years ago
The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-
WINSTONCH [101]

Answer: 1.212(10)^{-10} m

Explanation:

The de Broglie wavelength \lambda is given by the following formula:

\lambda=\frac{h}{p} (1)

Where:

h=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

p is the momentum of the atom, which is given by:

p=m_{e}v (2)

Where:

m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg is the mass of the electron

v=6(10)^{6}m/s is the velocity of the electron

This means equation (2) can be written as:

p=(9.11(10)^{-31}kg)(6(10)^{6}m/s) (3)

Substituting (3) in (1):

\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)} (4)

Now, we only have to find \lambda:

\lambda=1.2122(10)^{-10} m>>> This is the de Broglie wavelength of the electron

8 0
3 years ago
What is the speed of a wave with a period of 25 sec/cycle and a wavelength of 15 m/cycle?please help me !
Nuetrik [128]
The answer of this question is 0.6m/s
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(True or False Statement)
sertanlavr [38]
False is correct......
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