Answer:
Jesseca wanted to create a material that reflected most of the light that fell on it.
Explanation: The Graphite was the material in the passage that had reflected most of the light.
Answer:
conservative
Explanation:
Nonconservative force is the force that depends on a path, however conservative does not depend on a path and it is not associated with the potential energy. When the work is done by an unconservative force, mechanical energy is added or removed. Friction is the best example for a non-conservative force. When these non-conservative forces are acting, the mechanical energy changes but these are not preserved.
hope this helped!
Answer:
time rising = 34 / 9.8 = 3.47 sec
total time in air = 2 * 3.47 sec = 6.94 sec
(time rising must equal time falling)
R = 17 m/s * 6.94 s = 118 m
Can also use range formula
R = v^2 sin (2 theta) / g
tan theta = 34 / 17 = 2
theta = 63.4 deg
2 theta = 126.9 deg
sin 126.9 = .8
v^2 = 17^2 + 34^2 = 1445 m^2/s^2
R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above
Answer:
The acceleration of the earth is 7.05 * 10^-25 m/s²
Explanation:
<u>Step 1:</u> Data given
mass of the apple = 0.43 kg
acceleration = 9.8 m/s²
mass of earth = 5.98 * 10 ^24 kg
<u>Step 2:</u> Calculate the acceleration of the earth
Following the third law of Newton F = m*a
F(apple) = F(earth) = m(apple)*a(apple)
F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N
a(earth) = F(apple/earth)/m(earth)
a(earth) = 4.214N /5.98 * 10 ^24 kg
a(earth) = 7.05 * 10^-25 m/s²
The acceleration of the earth is 7.05 * 10^-25 m/s²
Answer:
17.1
Explanation:
The distance ahead, of the deer when it is sighted by the park ranger, d = 20 m
The initial speed with which the ranger was driving, u = 11.4 m/s
The acceleration rate with which the ranger slows down, a = (-)3.80 m/s² (For a vehicle slowing down, the acceleration is negative)
The distance required for the ranger to come to rest, s = Required
The kinematic equation of motion that can be used to find the distance the ranger's vehicle travels before coming to rest (the distance 's'), is given as follows;
v² = u² + 2·a·s
∴ s = (v² - u²)/(2·a)
Where;
v = The final velocity = 0 m/s (the vehicle comes to rest (stops))
Plugging in the values for 'v', 'u', and 'a', gives;
s = (0² - 11.4²)/(2 × -3.8) = 17.1
The distance the required for the ranger's vehicle to com to rest, s = 17.1 (meters).
