Answer:
<h2>
44 m/s</h2>
Explanation:
In this problem we are expected to calculate the velocity of Georges movements.
Given data
Total distance covered by George= 850+250= 1100 meters
Time taken by George to cover the total distance= 25 seconds
We know that velocity is, v= distance/ time
Therefore substituting our data into the expression for velocity we have
v= 1100/ 25= 44 m/s
Hence the velocity in m/s is 44
Answer:
If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.
The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m
s = ut + 1 / 2 at²
s = Distance
u = Initial velocity
t = Time
a = Acceleration
Vertically,
s = 15.4 m
u = 0
a = 9.8 m / s²
15.4 = 0 + ( 1 / 2 * 9.8 * t² )
t² = 3.14
t = 1.77 s
Horizontally,
u = 3.01 m / s
a = 0 ( Since there is no external force )
s = ( 3.01 * 1.77 ) + 0
s = 5.34 m
Therefore, the distance travelled by the ball before hitting the ground is 5.34 m
To know more about distance travelled
brainly.com/question/12696792
#SPJ1
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the coefficient of friction is 0.32
Explanation:
Given the data in the question;
we make use of kinematic equation of motion;
ω = ω₀ + ∝t
we substitute
ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )
ω = 3.9612 rad/s
The centripetal force acting on the sample is;
Fc = mrω²
from the image; r = 200 mm = 0.2 m
so we substitute
Fc = m(0.2 m ) ( 3.9612 rad/s )²
Fc = (3.13822 m/s²)m
we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;
f = Fc
μN = Fc
μmg = (3.13822 m/s²)m
μ = (3.13822 m/s²)m / mg
μ = (3.13822 m/s²) / g
acceleration due to gravity g = 9.8 m/s²
so
μ = (3.13822 m/s²) / 9.8 m/s²
μ = 0.32
Therefore, the coefficient of friction is 0.32