Answer: F(t) = 11 - 0.9(t)
Explanation:
We know the following:
The candle burns at a ratio given by:
Burning Ratio (Br) = 0.9 inches / hour
The candle is 11 inches long.
To be able to create a function that give us how much on the candle remains after turning it after a time (t). We will need to know how much of the candle have been burned after t.
Let look the following equation:
Br = Candle Inches (D) / Time for the Candle to burn (T) (1)
Where (1) is similar to the Velocity equation:
Velocity (V) = Distance (D)/Time(T)
This because is only a relation between a magnitude and time.
Let search for D on (1)
D = Br*T (2)
Where D is how much candle has been burn in a specif time
To create a function that will tell us how longer remains of the candle after be given a variable time (t) we use the total lenght minus (2):
How much candle remains? ( F(t) ) = 11 inches - Br*t
F(t) = 11 - 0.9(t)
F(t) defines the remaining length of the candle t hours after being lit
The kinetic energy of an object of mass m moving with speed v is given by:
![K= \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20)
For the bicycle in our problem,
![m=10 kg](https://tex.z-dn.net/?f=m%3D10%20kg)
and
![v=2 m/s](https://tex.z-dn.net/?f=v%3D2%20m%2Fs)
, so the kinetic energy is
Answer:
1.38*10^18 kg
Explanation:
According to the Newton's law of universal gravitation:
![F=G*\frac{m_a*m_p}{r^2}](https://tex.z-dn.net/?f=F%3DG%2A%5Cfrac%7Bm_a%2Am_p%7D%7Br%5E2%7D)
where:
G= Gravitational constant (6.674×10−11 N · (m/kg)2)
ma= mass of the astronaut
mp= mass of the planet
![F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2](https://tex.z-dn.net/?f=F%3Dm_a.a%5C%5C%28v_f%20%29%5E2%3D%28v_o%29%5E2%2B2.a.%5CDelta%20y%5C%5C%5C%5Ca%3D%5Cfrac%7B%28v_f%29%5E2-%28v_o%29%5E2%7D%7B2.%5CDelta%20y%7D%5C%5C%5C%5Ca%3D%5Cfrac%7B%280%29%5E2-%284.29m%2Fs%29%5E2%7D%7B2.0.64m%7D%3D14.38m%2Fs%5E2%5C%5C%5C%5CF%3Dm_a%2A14.38m%2Fs%5E2)
so:
![m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg](https://tex.z-dn.net/?f=m_a%2A14.38m%2Fs%5E2%3D%286.674%2A10%5E%7B-11%7DN.%28m%2Fkg%29%5E2%29%2A%5Cfrac%7Bm_a.m_p%7D%7B%282.530%2A10%5E3m%29%5E2%7D%5C%5Cm_p%3D%5Cfrac%7B14.38m%2Fs%5E2%282.530%2A10%5E3m%29%5E2%7D%7B%286.674%2A10%5E%7B-11%7DN.%28m%2Fkg%29%5E2%29%7D%5C%5C%5C%5Cm_p%3D1.38%2A10%5E%7B18%7Dkg)
Answer:what are the answer options?
Explanation
The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V
<u>Explanation:</u>
Given data,
E= 3 ×10 ⁶ Δx=0.06/100
We have to find the minimum potential difference
E= -ΔV/Δx
ΔV=- E × Δx
ΔV =-3 ×10 ⁶ . 0.06/100
ΔV=-1800 V
The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V