The greater the mass the greater is inertia.
Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:
![V=\frac{\Delta W}{\Delta q}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B%5CDelta%20W%7D%7B%5CDelta%20q%7D)
Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq
![I=\frac{\Delta q}{\Delta t}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B%5CDelta%20q%7D%7B%5CDelta%20t%7D)
Where:
I is the current (2 A)
Δt is the time (20 s)
![2=\frac{\Delta q}{20}](https://tex.z-dn.net/?f=2%3D%5Cfrac%7B%5CDelta%20q%7D%7B20%7D)
![q=40 C](https://tex.z-dn.net/?f=q%3D40%20C)
Then, we can put this value of charge in the voltage equation.
![V=\frac{4600}{40}=115 V](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4600%7D%7B40%7D%3D115%20V)
Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
It mainly travels by kinetic energy
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
![9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}](https://tex.z-dn.net/?f=9.8%5Ctimes%2012.5%3DW%5Ctimes%20%2850-12.5%29%5C%5C%5C%5CW%3D%5Cdfrac%7B9.8%5Ctimes%2012.5%7D%7B37.5%7D)
W = 3.266 N
The mass of the meters stick is :
![m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BW%7D%7Bg%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B3.266%7D%7B9.81%7D%5C%5C%5C%5Cm%3D0.333%5C%20kg)
So, the mass of the meter stick is 0.333 kg.
Answer:
![t_1 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=%20t_1%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
![t_2 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=%20t_2%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
Δd = ![v_it_1 = v_i^2/a_i](https://tex.z-dn.net/?f=%20v_it_1%20%3D%20v_i%5E2%2Fa_i)
Explanation:
As
, when the car is making full stop,
.
. Therefore,
![0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=0%20%3D%20v_i%20-%20a_it_1%5C%5Cv_i%20%3D%20a_it_1%5C%5Ct_1%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
Apply the same formula above, with
and
, and the car is starting from 0 speed, we have
![v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=%20v_i%20%3D%200%20%2B%20a_it_2%5C%5Ct_2%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
As
. After
, the car would have traveled a distance of
![s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}](https://tex.z-dn.net/?f=s%28t%29%20%3D%20s%28t_1%29%20%2B%20s%28t_2%29%5C%5Cs%28t_1%29%20%3D%20%28v_it_1%20-%20%5Cfrac%7Ba_it_1%5E2%7D%7B2%7D%29%5C%5C%20s%28t_2%29%20%3D%20%5Cfrac%7Ba_it_2%5E2%7D%7B2%7D)
Hence ![s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}](https://tex.z-dn.net/?f=%20s%28t%29%20%3D%20%28v_it_1%20-%20%5Cfrac%7Ba_it_1%5E2%7D%7B2%7D%29%20%2B%20%5Cfrac%7Ba_it_2%5E2%7D%7B2%7D%20)
As
we can simplify ![s(t) = v_it_1](https://tex.z-dn.net/?f=s%28t%29%20%3D%20v_it_1)
After t time, the train would have traveled a distance of ![s(t) = v_i(t_1 + t_2) = 2v_it_1](https://tex.z-dn.net/?f=%20s%28t%29%20%3D%20v_i%28t_1%20%2B%20t_2%29%20%3D%202v_it_1)
Therefore, Δd would be ![2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i](https://tex.z-dn.net/?f=%202v_it_1%20-%20v_it_1%20%3D%20v_it_1%20%3D%20v_i%5E2%2Fa_i)