If an object is moving eastward and slowing down, then the direction of its acceleration vector is

1 answer:

8 0

It would still be eastward because The direction of the velocity vector is always in the same direction as the direction which the object moves.

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In a previous problem, you calculated how much energy was needed to accelerate a 2,100-kg Tesla Model S from 0 to 100 km/hr, and

Naddika [18.5K]

Answer:

t=2.348s

Explanation:

First we need to calculate the Potential Energy of a Tesla to reach the 100km/h

So we know that,

$W=KE_f - KE_i = \frac{1}{2}m(V_f^2-V_i^2)=\frac{1}{2}(2100)(27.78^2)$ $W=8.1*10^5J$

Here we can to calculate the time through,

$t=\frac{W}{P}$

$t=\frac{8.1*10^5}{345*10^3}$

$t=2.348s$

5 0

3 years ago

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res

murzikaleks [220]

Answer:

(a) $V=11.86\ V$

(b) $V=9.76\ V$

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

$V=R.I$

(a) The electromagnetic force of the battery is $\varepsilon =12.6\ V$ and its internal resistance is $R_i=0.06\ \Omega$ . Knowing the equivalent resistance of the headlights is $R_e=5.2\ \Omega$ , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

$\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)$

Solving for i

$\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A$

i=2.28\ A

The potential difference across the headlight bulbs is

$V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V$

$V=11.86\ V$

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

$V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V$