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Step2247 [10]
3 years ago
8

Dumbo was picked on for having big. what might be an advantage to havin large ears

Physics
1 answer:
Aleonysh [2.5K]3 years ago
7 0

Answer:

Well the reason that elephants have big ears in the first place is to cool off. They flap their ears and use them as fans.

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In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find
katrin2010 [14]

Answer:

a = 1.055 x 10¹⁷ m/s²

Explanation:

First, we will find the force on electron:

E = \frac{F}{q}\\\\F = Eq\\

where,

F = Force = ?

E = Electric Field = 6 x 10⁵ N/C

q = charge on electron = 1.6 x 10⁻¹⁹ C

Therefore,

F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\

F = 9.6 x 10⁻¹⁴ N

Now, we will calculate the acceleration using Newton's Second Law:

F = ma\\a = \frac{F}{m}\\

where,

a = acceleration = ?

m = mass of electron = 9.1 x 10⁻³¹ kg

therefore,

a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\

<u>a = 1.055 x 10¹⁷ m/s²</u>

5 0
2 years ago
Oil _____.
atroni [7]
Oil <span>must be mined from underground 

Oil is NOT a renewable resource
Oil does NOT release more toxins than coal
Oil also is NOT easily replenished

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4 0
3 years ago
Read 2 more answers
In terms of volume,how do ml &amp; cm3 relate to one another?
Keith_Richards [23]

1 milliliter = 1 cubic centimeter (cm^3)

3 0
3 years ago
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A city bus travels along its route from Jackson Street 8 km to Aurora Avenue, where it turns and continues 13 km.
Elina [12.6K]

im a bit confused on what the question is:(

if its just asking for the total number of km traveled it would be 21 km

3 0
2 years ago
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
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