Answer:
The resulting velocity of the ball after it hits the racket was of V= 51.6 m/s
Explanation:
m= 55.6 g = 0.0556 kg
t= 2.8 ms = 2.8 * 10⁻³ s
F= 1290 N/ms * t - 330 N/ms² * t²
F= 1024.8 N
F*t= m * V
V= F*t/m
V= 51.6 m/s
Answer:
B) Friction
Explanation:
Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.
For an object sliding on a flat surface, the force of friction has magnitude:
where
is the coefficient of kinetic friction
m is the mass of the object
g is the acceleration of gravity
The direction of the force of friction is always opposite to the direction of motion of the object.
In reality, friction also acts if the object is at rest and it is pushed by a force; in this case, we talk about static friction, and its magnitude is
where 
is called coefficient of static friction, and it is generally larger than the coefficient of kinetic friction.
Answer:
Momentum, p = 23250 kg m/s
Explanation:
Given that
Mass of a car, m = 1550 kg
Speed pf car, v = 15 m/s
We need to find the momentum of the car. The formula for the momentum of an object is given by :
p = mv
Substituting all the values in the above formula
p = 1550 kg × 15 m/s
p = 23250 kg m/s
So, the momentum of the car is 23250 kg m/s.
D. all of these
all of these use electricity
Hope I helped!
consider the motion along the horizontal direction :
v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s
X = horizontal displacement of the ball = 2.0 m
a = acceleration along the horizontal direction = 0 m/s²
t = time taken to land = ?
using the kinematics equation
X = v₀ t + (0.5) a t²
2.0 = 3.0 t + (0.5) (0) t²
t = 2/3
consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s
Y = vertical displacement of the ball = height of the table = h
a = acceleration along the vertical direction = 9.8 m/s²
t = time taken to land = 2/3
using the kinematics equation
Y = v₀ t + (0.5) a t²
h = 0 t + (0.5) (9.8) (2/3)²
h = 2.2 m
C 2.2 m
