Answer:
The net force on the skater is zero. (
)
Explanation:
According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. (
)
If the kinetic energy of each ball is equal to that of the other,
then
(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²
Multiply each side by 2:
(mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²
Divide each side by (mass of gb) and by (speed of ppb)² :
(mass of ppb)/(mass of gb) = (speed of gb)²/(speed of ppb)²
Take square root of each side:
√ (ratio of their masses) = ( 1 / ratio of their speeds)²
By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written. Let's just come in from the
cold, and say it the clear, easy way:
If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.
If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.
The pingpong ball is moving faster than the golf ball.
The directions of their motions are irrelevant.
Answer:
0.61°
Explanation:
Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.
Pulling force= resistance force
From the formula for pulling force,
F(x)= Fcos(θ)
= 425×cos(35.2)
=347N
The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)
=425×0.567=245N
Resistance force= (325N+ 245N) (α)= 570N(α)
We can now equates the pulling force to resistance force
570 (α)= 347N
(α)= 347/570
= 0.61
They have similar physical and chemical properties because of thier valence electrons
Answer
given,
y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]
length of the rope = 1.33 m
mass of the rope = 3.31 g
comparing the given equation from the general wave equation
y(x,t)= A cos[k x+ω t]
A is amplitude
now on comparing
a) Amplitude = 2.20 mm
b) frequency =


f = 118.25 Hz
c) wavelength




d) speed


v = 105.84 m/s
e) direction of the motion will be in negative x-direction
f) tension


T = 27.87 N
g) Power transmitted by the wave


P = 0.438 W