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Anton [14]
3 years ago
7

Jane's brother Andrew leaves home for school at 8:00 am. He walks at 3.3 kph. At 8:20 am Jane discovers that Andrew has left his

homework at home. She decides to follow him on her bicycle and give him his homework, but she wants to be back home in time for an online test at 9 am. What is the minimum constant speed at which she needs to ride? (Hint: At what time of day will they meet?)
Jane must ride at least ______ kph.
Physics
1 answer:
Gre4nikov [31]3 years ago
7 0
8;30
i think hope i helped
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Vitamins and iron are nutrients that provide energy to the human body.<br> A true<br> B false
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I think it is A. but then you can also produce your own energy
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If Fg=mg solve for g
Bas_tet [7]

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

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Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

or

T =  \frac{1}{1.03\times 10^8}  = 9.6 x 10⁻⁹ seconds  

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

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6 0
3 years ago
Globalization is the process of
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Read 2 more answers
A disk-shaped merry-go-round of radius 2.63 m and mass 152 kg rotates freely with an angular speed of 0.526 rev/s. A 51.7 kg per
zalisa [80]

 Explanation:

Given

radiusr=2.63 m

N=0.526 rev/s

\omega =3.30 rad/s

mass disc  M=152 kg

mass of person  m=51.7 kg

velocity of Person  v=2.76 m/s

moment of inertia  I=Mr^2

I=0.5\times 152\times 2.63^2=827.64 kg-m^2

Initial angular momentum

L_i=I\omega +mvr

L_i=827.64\times 3.30+51.7\times 2.76\times 2.63

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Final Moment inertia

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Conserving angular momentum

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\omega _f=2.62 rad/s

4 0
3 years ago
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