Answer:
a) 12.8212 N
b) 12.642 N
Explanation:
Mass of bucket = m = 0.54 kg
Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s
Speed of sand = 3.2 m/s
g= 9.8 m/sec2
<u>Condition (a);</u>
Mass of sand = Ms = 0.75 kg
So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg
== > total weight = 1.29 × 9.8 = 12.642 N
Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N
Scale reading is sum of impact of sand and weight force ;
i-e
scale reading = 12.642 N+0.1792 N = 12.8212 N
<u>Codition (b);</u>
bucket mass + sand mass = 0.54 +0.75=1.29 kg
==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)
The answer is polarization. It is just like when you are on a boat and use polarized glasses so the water wont reflect into your eyes it also intensifies the color of the water so you can see better into the water.
example: you cant see any fish because glare from the sun, put on polarized glasses and now you can see fish up to 30 feet deep.
The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.
<u>Given the following data:</u>
- Angle of inclination = 25°
To determine the X and Y components of the force:
<h3>The horizontal component (X) of a force:</h3>
Mathematically, the horizontal component of a force is given by this formula:
Fx = 90.63 Newton.
<h3>The vertical component (Y) of tensional force:</h3>
Mathematically, the vertical component of a force is given by this formula:
Fy = 42.26 Newton.
Read more on horizontal component here: brainly.com/question/4080400
Answer:
Explanation:
It would actually be A. 30 , as each hour of ascension (i am not sure about the correct terminology) equals 15 .
Answer:
The transverse wave will travel with a speed of 25.5 m/s along the cable.
Explanation:
let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.
then, if V is the volume of the cable:
ρ = m/V
m = ρ×V
but V = A×L , where L is the length of the cable.
m = ρ×(A×L)
m/L = ρ×A
then the speed of the wave in the cable is given by:
v = √(T×L/m)
= √(T/A×ρ)
= √[2.96×10^4/(4.49×10^-3×7860)]
= 25.5 m/s
Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.
