Answer:
Gravity changes with altitude. as we know The gravitational force is proportional to 1/R2, where R is your distance from the center of the Earth.
eg. The radius of the Earth at the equator is 6400 kilometers.
Let's say you were in a jet at the equator that was 40 kilometers high above the earth's surface.
may be helpfull
<em>Convert 1nanosecond in to its SI init</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>
Answer:
-2.79 m/s²
Explanation:
Given:
v₀ = 20 m/s
v = 11 m/s
Δx = 50 m
Find: a
v² = v₀² + 2aΔx
(11 m/s)² = (20 m/s)² + 2a (50 m)
a = -2.79 m/s²
Round as needed.
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
<h2>The temperature of the air is 66.8° C</h2>
Explanation:
From the Newton's velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .
In this case The velocity of sound = frequency x wavelength
= 798 x 0.48 = 383 m/sec
Suppose the temperature at this time = T K
Thus 383 ∝ 
I
The velocity of sound is 329 m/s at 273 K ( given )
Thus 329 ∝ 
II
Dividing I by II , we have
= 
or 
= 1.25
and T = 339.8 K = 66.8° C
