Answer:
995.12 N/C
Explanation:
R = 9 cm = 0.09 m
σ = 9 nC/m^2 = 9 x 10^-9 C/m^2
r = 9.1 cm = 0.091 m
q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C
E = kq / r^2
E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)
E = 995.12 N/C
Answer:
a. wavelength of the sound, 
b. observed frequecy, 
Given:
speed of sound source, 
= 80 m/s
speed of sound in air or vacuum, 
= 343 m/s
speed of sound observed, 
= 0 m/s
Solution:
From the relation:
v = 
(1)
where
v = velocity of sound
= observed frequency of sound
= wavelength
(a) The wavelength of the sound between source and the listener is given by:
(2)
(b) The observed frequency is given by:
(3)
Using eqn (2) and (3):
We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers.
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).
Earth's radius . . . . . 6,360 km = 6.36 x 10⁶ meters
Moon's radius . . . . . 1,738 km = 1.738 x 10⁶ meters
Sum of their radii = 8.098 x 10⁶ meters
Also:
Earth's mass . . . . . 5.972 x 10²⁴ kg
Moon's mass . . . . . 7.348 x 10²² kg
<span>
and now we're ready to go !
Gravitational force =
G M₁ M₂ / R²
= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²² kg)/</span>(8.098 x 10⁶ m)²
= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³) Newtons
= (I get ...) 4.463 x 10²³ Newtons
That's almost exactly 10²³ pounds
= 50,153,000,000,000,000,000 tons.
Those are big numbers.
All I can say is: I wouldn't exactly call that "resting" on the surface".
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>
