The coefficient of friction between the road and the car's tire is determined as 0.78.
<h3>Acceleration of the car</h3>
The acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
- u is the initial velocity = 97 km/h = 26.94 m/s
a = (26.94)²/(2 x 47)
a = 7.72 m/s²
<h3>Coefficient of friction</h3>
μ = a/g
μ = (7.72)/9.8
μ = 0.78
Learn more about coefficient of friction here: brainly.com/question/14121363
#SPJ1
Answer:
the correct one is D,
Ultraviolet, x-ray, gamma ray
Explanation:
Electromagnetism radiation are waves of energy that is expressed by the Planck relationship
E = h f
where h is the plank constant and f the frequency of the radiation.
Also the speed of light is
c = λ f
we substitute
E = h c /λ
therefore to damage the cells of the body radiation of appreciable energy is needed
microwave radiation has an energy of 10⁻⁵ eV
infrared radiation E = 10⁻² eV
visible radiation E = 1 to 3 eV
radiation Uv E = 3 to 6 eV
X-ray E = 10 eV
gamma rays E = 10 5 eV
therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.
When checking the answers, the correct one is D
0.77 m/s2 directed 35° south of west
net force = (-17,-12)
net force = mass * acceleration
(-17,-12) = 27 * (x-acceleration,y-acceleration)
(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)
angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.
magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)
Answer:
aw why? are you deleting the app for school?
Analog
Television transmitters use one of two different technologies: analog, in which the picture and sound are transmitted by analog signals modulated onto the radio carrier wave, and digital in which the picture and sound are transmitted by digital signals.
