Answer:
This means C.the car uses 20% of the energy store in the gasoline for motion.
Explanation:
I hope this helps.
Answer:
As 28m/s = 28m/s
Explanation:
r = the radius of the curve
m = the mass of the car
μ = the coefficient of kinetic friction
N = normal reaction
When rounding the curve, the centripetal acceleration is

therefore



As 28m/s = 28m/s
Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
Radio waves in a vacuum travel at the speed of light because they are a type of electromagnetic radiation like a light has been measured as traveling at 3×10^8 m/s in a vacuum.
Charged particles that are accelerating, like time-varying electric currents, are what produce radio waves. Radio and television signals are transmitted using radio waves, and microwaves used in radar and microwave ovens are also radio waves. Radio waves are emitted by a lot of celestial bodies, including pulsars. High RF exposure levels have the potential to heat biological tissue and raise body temperature. The body's inability to handle or remove the extra heat that could be generated by high RF exposure in humans could result in tissue damage.
To learn more about radio waves please visit -
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A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).
We can calculate the initial velocity of the car (u) using the following kinematic equation.
![v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5C%5C%5Cu%20%3D%20%5Csqrt%5B%5D%7Bv%5E%7B2%7D-2as%7D%20%3D%20%5Csqrt%5B%5D%7B%280m%2Fs%29%5E%7B2%7D-2%28-42.61m%2Fs%5E%7B2%7D%20%29%2810.99m%29%7D%20%3D%2030.60m%2Fs)
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
Learn more: brainly.com/question/14851168