Answer:
The car's initial velocity is 195.96 m/s
Explanation:
Given;
the length of the mark, x = 9.6 km = 9600 m
acceleration of the car, a = -2 m/s²
Apply the kinematic equation below
v² = u² + 2ax
where;
v is the final velocity = 0
u is the initial velocity
simplify the equation and you will obtain;
0 = u² + 2ax
0 = u² + 2(-2)(9600)
0 = u² - 38400
u² = 38400
u = √38400
u = 195.96 m/s
Therefore, the car's initial velocity is 195.96 m/s
FIRST STOP: EARTH'S ATMOSPHERE
Once the sun's energy reaches earth, it is intercepted first by the atmosphere. A small part of the sun's energy is directly absorbed, particularly by certain gases such as ozone and water vapor. Some of the sun's energy is reflected back to space by clouds and the earth's surface.
Explanation:
36-4/4= 9 m/squared. meter per squared.
acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)
Answer:
The string must support the tension of 392 N.
Explanation:
The tension that the string must support should equal the centripetal force exerted on the on the stone as it goes in a circular path (because if the string supported less tension, it would break).
The centripetal force exerted on the stone is
where
<em>v</em> = velocity of the stone in m/s
<em>m</em> = mass of the stone in kg
<em>R</em> = radius of the circular path.
Now the velocity of the stone is 7.00 m/s, the mass of the stone is 4000g or 4 kg (1000 g = 1kg), and the radius of the circular path is just the length of the string, and it is 50 cm or 0.5 m (100cm =1m); therefore, we get
m = 4kg
v =7m/s
R = 0.5m.
We put these values into the equation for the centripetal force and get:
The centripetal force is 392 Newtons, and therefore, the tension that the string must support mus be 392 N.
Answer:
Option (B) is true
Explanation:
refractive index of water = n1
refractive index of air = na = 1
refractive index of oil = n2
When the ray goes from water to air
Use Snell's law
Let the angle of incidence is i
n1 Sin i = na x Sin r
For total internal reflection, r = 90°
n1 x Sin i = 1 x Sin 90
Sin i = 1 / n1 .... (1)
For water oil interface
angle of incidence is i and let the angle of refraction is r.
n1 x Sin i = n2 x Sin r
n1 x / 1 n1 = n2 Sin r (from equation (1)
Sinr = 1/n2