Answer:
Explanation:
For rolling up or down an incline plane , the acceleration or deceleration of the rolling body is given by the following expression
a = g sinθ / (1 + k²/r² )
where k is radius of gyration of rolling body and θ is angle of inclination
a = g sin15 / ( 1 + 1 ) [ for hoop k = r ]
a = 9.8 x .2588 / 2
= 1.268 m / s²
a )
Let s be the distance up to which it goes
v² = u² - 2as
0 = 3.3² - 2 x 1.268 s
s = 4.3 m
b ) Let time in going up be t₁
v = u - at₁
0 = 3.3 - 1.268 t₁
t₁ = 2.6 s
Time in going down t₂
s = 1/2 a t₂²
4.3 = .5 x 1.268 t₂²
t₂ = 2.60
Total time
= t₁ +t₂
= 2.6 + 2.6
= 5.2 s
Answer:
c
Explanation:
this is because you move ( kinetic energy) your hands and clap the other person's hand, you will definitely hear sound right.
I think it's Barium sulfate, the soild and percipitate
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
Answer:
The angle above the horizontal at which the pitcher throws the ball determines the distance the ball travels before returning to the height at which it was thrown
Explanation:
The baseball is thrown as a projectile and the range, 'R', of the baseball which is the distance the baseball travels before the height above the ground returns to the initial height is given given as follows;
Where;
R = The range of the baseball = The horizontal distance away from the pitcher the ball reaches
u = The initial velocity with which the baseball was thrown
θ = The angle above horizontal a baseball pitcher throws the ball
g = The acceleration due to gravity ≈ 9.81 m/s²
From the the equation, when θ = 0, sin(θ) = sin(0) = 0 and the ball does not cover any horizontal distance before going lower than the height at which it was thrown, therefore, for the ball to travel further, the angle of launch, θ has to be larger than 0.
