The answer is D. Products are formed from reactants by the breaking and forming of new bonds.
Answer:
9.34 N
Explanation:
First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:
where
f is the frequency of the wave
is the wavelength
For the waves in this string we have:
, since it completes 625 cycles per second
is the wavelength
So the speed of the wave is
The speed of the waves in a string is related to the tension in the string by
(1)
where
T is the tension in the string
is the linear density
In this problem:
is the mass of the string
L = 0.75 m is the its length
Solving the equation (1) for T, we find the tension:
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
Answer:N=322.53 rpm
Explanation:
Given
Linear velocity (v)=1.25 m/s
Position from center is 3.7 cm
we know
and
N=322.53 rpm
Answer:
0.243 m/s
Explanation:
From law of conservation of motion,
mu+m'u' = V(m+m')................. Equation 1
Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.
make V the subject of the equation
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s
Substitute into equation 2
V = (260000×0.32+52500×(-0.14))/(260000+52500)
V = (83200-7350)/312500
V = 75850/312500
V = 0.243 m/s