Answer:
Explanation:
Given
Both cars mass is m
and solving problem in Vertical and horizontal direction
considering + y and +x to be positive and u be the final velocity of system
Conserving Momentum in Vertical direction

------1
Conserving momentum in x direction
-----2
squaring and adding 1 &2




Explanation:
To convert from scientific notation to decimal, move the decimal point 5 places to the left.
5.93×10⁻⁵ = 0.0000593
This lab is investigating the relationship between mass, <u>Speed </u>, and momentum.
Momentum is manufactured from the mass and speed of an object. it's miles a vector quantity, owning a significance and a direction. If m is an object's mass and v is its speed, then the object's momentum is p.
Momentum in an easy way is a quantity of movement. right here amount is measurable because if an item is moving and has mass, then it has momentum. If an object no longer flows then it has no momentum. however, in regular existence, it has an important but many people didn't understand it.
Momentum gives the connection between the mass, pace, and direction of an object. Any exchange in momentum results in pressure. So, an exchange in momentum is used to determine the force appearing upon the item.
Learn more about momentum here:-brainly.com/question/1042017
#SPJ1
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
Answer:
yes driver exceed the car speed
Explanation:
given data
speed of car = 38 m/s
speed limit = 75.0 mi/h
to find out
Is the driver exceeding the speed limit
solution
we know car speed is 38 m/s and limit is 75 mi/hr
so for compare the speed limit we convert limit and make them same
as we know
1 m/s = 2.236 mi/hr
so
car speed 38 m/s = 38 × 2.236 = 85.003 mi/hr
as this car speed is exceed the speed limit that is 75 mi/hr
yes driver exceed the car speed