To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.
PART A)
The time taken to travel a distance of 250km with a speed of 95km/h is
Time taken for the lunch is
The time taken travel a distance of 250km with a speed of 55km/h
The total time taken is
The average speed is the ratio of total distance and total time
PART B)
As the displacement is zero the average velocity is zero.
Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
Answer:
The induced current is 0.084 A
Explanation:
the area given by the exercise is
A = 200 cm^2 = 200x10^-4 m^2
R = 5 Ω
N = 7 turns
The formula of the emf induced according to Faraday's law is equal to:
ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt
Replacing values:
ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V
the induced current is equal to:
I = ε /R = 0.42/5 = 0.084 A
The energy transferred to the spring is given by:
where
k is the spring constant
x is the elongation of the spring with respect its initial length
Let's convert the data into the SI units:
so now we can use these data inside the equation ,to find the energy transferred to the spring:
To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.
The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was
The rate of energy loss per day would then be,
That is to say that Energy in Jules per lost day is 5356800J
By definition we know that 
In this way the energy in Cal is,
The number of kilocalories (food calories) must be 1279.694 KCal
