tje average acceleration of the car is
the average acceleration of the car is 1.4 m/s
The radius of the Earth is about 3,960 miles.
That's how far the center of the Earth is straight down under your feet.
It's farther than the distance all the way across the USA, from New York
to Los Angeles. It's like the distance from London (England) to Chicago.
If you drill absolutely straight and aim straight down, that's how far
you need to drill to reach the center. Unfortunately, your drill will melt
long before you get there.
The longest hole ever drilled into the Earth (so far) is the Sakhalin-I Oduptu Well,
drilled by Russia into the sea-bottom in 2011.
That hole is 40,502 feet ... roughly 7.67 miles, or about 0.0019 of the distance
to the center of the Earth. You can see that we still have quite a long way to go !
Answer:
2200000 = 2.2E6 min for light from Proxima to reach earth
8.3 min from light sun to reach earth
2.2E6/8.3 = 2.56E5 times for light from Proxima
Proxima is about 256,000 times farther away than the sun
Since the sun is about 93,000,000 = 9.3E7 miles from earth
Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away
Note - the speed of light is
3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given
Answer:
b. 600,000 J
Explanation:
Applying the law of conservation of energy,
The thermal energy created = Kinetic energy of the suv.
Q' = 1/2(mv²)............... Equation 1
Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.
From the question,
Given: m = 3000 kg, v = 20 m/s
Substitute these values into equation 1
Q' = 1/2(3000×20²)
Q' = 600000 J
Hence the right option is b. 600,000 J
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
