Answer:
none
Explanation:
~both of them show to the nearest metre.
~millimeter has (mm) unit eg 0.7mm
Answer:
<h3> b. 1.18</h3>
Explanation:
The fundamental frequency in string is expressed as;
F1 = 1/2L√T/m .... 1
L is the length of the string
T is the tension
m is the mass per unit length
If the tension is increased by 40%, the new tension will be;
T2 = T + 40%T
T2 = T + 0.4T
T2 = 1.4T
The new fundamental frequency will be;
F2 = 1/2L√1.4T/m ..... 2
Divide 1 by 2;
F2/F = (1/2L√1.4T/m)/1/2L√T/m)+
F2/F = √1.4T/m ÷ √T/m
F2/F = √1.4T/√m ×√m/√T
F2/F = √1.4T/√T
F2/F = 1.18√T/√T
F2/F = 1.18
F2 = 1.18F
Hence the fundamental frequency of vibration changes by a factor of 1.18
Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = 
-K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W -
= -K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3
]
v = √(31.63 2/4.3)
v = 3.84 m/s
I’m going to use molasses as an example of a substance.
The mass and volume both change when changing the amount of molasses.
However, the density does not change. This is because the mass and volume increase at the same rate/proportion!
Even though there is more molasses (mass) in test tube A, the molasses also takes up more space (volume). Therefore, the spacing between those tiny particles that make up the molasses is constant (does not change).
The size or amount of a material/substance does not affect its density.
