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larisa86 [58]
3 years ago
7

The Tesla Roadster electric car is rated at 16.8kW how much energy does it use during a one hour car drive

Physics
1 answer:
Damm [24]3 years ago
7 0

Answer: 6.05\cdot 10^7 J

Explanation:

Energy is related to power by:

E=Pt

where P is the power and t is the time. In this problem, we have:

P=16.8 kW=16800 W

t=1 h=3600 s

So, the energy used in one hour is

E=(16800 W)(3600 s)=6.05\cdot 10^7 J

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Pls I need it fast , its for my homework and I can’t find it
frez [133]

Answer:

first one is  series second one is paralle

Explanation:

5 0
3 years ago
How do you find the speed of an object given its mass and kinetic energy (what is the formula)?
madam [21]
   v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ; 
        
and solve for "v".

______________________________________________________
Explanation:
_____________________________________________________
The formula is:  KE = (½) * (m) * (v²) ;
_____________________________________
  
"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
Note:  Velocity is similar to speed, in that velocity means "speed and direction";  however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ;  to solve for "(v)" ; velocity, which is very similar to                                          the "speed"; 
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ;  ↔  (½)*(m)* (v²) = (KE) ; 
___________________________________________________

→ We have:  (½)*(m)* (v²) = (KE)  ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
   
→ We divide each side of the equation by: "[(½)* (m)]" ; 
___________________________________________________
    
           →   [ (½)*(m)*(v²) ] /  [(½)* (m)]  = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
 to get: 
______________________________________________________
                           →   v²     =   (KE) / [(½)* (m)]
                     
                           →   v²     = 2 KE / m
_______________________________________________________
Take the "square root" of each side of the equation ;
_______________________________________________________
                          →  √ (v²)  =  √ { 2*(KE) ] / m }
________________________________________________________

                          →     v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"]; 
       
and solve for "v".

______________________________________________________
8 0
3 years ago
Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp
maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = \dfrac{N}{C}m

c = Speed of light = m/s

Momentum is given by

\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s

The unit of MeV/c in SI fundamental units is kg m/s

5 0
2 years ago
You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in
FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, V=108\ cm^3

We know that the volume of the cylinder is given by :

V=\pi r^2 h

108=\pi r^2 h    

h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
3 years ago
Recall: earth applies _____________ on Earth or equivalently
Olin [163]

Answer:

gravitational attraction

4 0
3 years ago
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