Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
Answer:
Your answer will be 600meters
Answer:
Technician A says that this is the normal operation of the ETC self -test is the correct answer.
Explanation:
An engine control unit (ECU), also widely referred to as an engine control module (ECM), is a type of electronic control device that controls an internal combustion engine with a series of actuators to ensure maximum engine performance.
It achieves so by reading values from a multitude of sensors within the engine bay, translating data using multidimensional feedback maps (the so-called lookup tables) and modifying the actuators.
Mechanically fixed and dynamically regulated by mechanical and pneumatic means, air-fuel combination, ignition time, and idle speed were before ECUs.
As soon as the system gets battery voltage, after ignition is turned, the efi computer makes a self-test of all the actuators and sensors, included the ETC.
Answer:
f = 931.1 Hz
Explanation:
Given,
Mass of the wire, m = 0.325 g
Length of the stretch, L = 57.7 cm = 0.577 m
Tension in the wire, T = 650 N
Frequency for the first harmonic = ?
we know,
μ is the mass per unit length
μ = 0.325 x 10⁻³/ 0.577
μ = 0.563 x 10⁻³ Kg/m
now,
v = 1074.49 m/s
The wire is fixed at both ends. Nodes occur at fixed ends.
For First harmonic when there is a node at each end and the longest possible wavelength will have condition
λ=2 L
λ=2 x 0.577 = 1.154 m
we now,
v = f λ
f = 931.1 Hz
The frequency for first harmonic is equal to f = 931.1 Hz
Here we have to add the two measurements given in the question
The measurement values are given as 1.0090 cm and 0.02 cm.we have to add them on the basis of significant figure rules.
As per the addition rule in terms of significant figures
1-First we have to select the number of significant digits after the decimal point of each quantity.
2-Now we have to remember that during the addition ,the resultant of two quantities will follow the quantity having least number of significant figures after the decimal point.
3-Here we are considering the minimum number of significant figures after the decimal points not the minimum number of significant figures in case of multiplication and division
Now we have to add these two quantities as per the above rule-
1.0090 cm +0.02 cm
=1.0290 cm
Here the result will follow 0.02 which has minimum number of significant figures after the decimal points.
Hence we have to round off the number from 9 of 1.0290
As 9 is greater than 5 ,so he actual result will be 1.03 cm
