The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.
<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
Ball 4 because the higher the elevation is the greater the potential energy it has
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
1 mile. Is this a joke lol
Answer:
The change in temperature is 
Explanation:
From the question we are told that
The temperature coefficient is 
The resistance of the filament is mathematically represented as
![R = R_o [1 + \alpha \Delta T]](https://tex.z-dn.net/?f=R%20%20%3D%20%20R_o%20%5B1%20%2B%20%5Calpha%20%20%5CDelta%20T%5D)
Where 
is the initial resistance
Making the change in temperature the subject of the formula
![\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BR%7D%7BR_o%7D%20-%201%20%5D)
Now from ohm law
This implies that current varies inversely with current so
Substituting this we have
![\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BI_o%7D%7BI%7D%20-%201%20%5D)
From the question we are told that
Substituting this we have
![\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BI_o%7D%7B%5Cfrac%7BI_o%7D%7B8%7D%20%7D%20-%201%20%5D)
=> 
