Answers
Explanation:
it is equal to the speed (v) of a wave train in a medium divided by its frequency (f): λ = v/f. Waves of different wavelengths.
Answer:
It is measured in the direction of the wave. Description: Wavelength is the distance from one crest to another, or from one trough to another, of a wave (which may be an electromagnetic wave, a sound wave, or any other wave).
Related Questions
a right circular cylinder with a radius of 2.3cm and a height of 1.4cm has a total surface area of
Answers
Answer:
6.44pie
Explanation:
[tex]2\pirh[/tex]
a effort of 100n can raise a load of 2000n in a hydraulic press. calculate the cross-sectional area of a small piston in it. The cross-sectional area of a large piston is 4m^s
Answers
Answer:
[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]
Explanation:
The pressure on the pistons is given as;
Pressure = [tex]\frac{Force}{Area}[/tex]
So that,
Pressure on the small piston = [tex]\frac{F_{1} }{A_{1} }[/tex] and Pressure on the large piston = [tex]\frac{F_{2} }{A_{2} }[/tex]
Thus,
[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
Given that: [tex]F_{1}[/tex] = 100 N, [tex]F_{2}[/tex] = 2000 N, [tex]A_{2}[/tex] = 4 [tex]m^{2}[/tex].
[tex]\frac{100}{A_{1} }[/tex] = [tex]\frac{2000}{4}[/tex]
[tex]A_{1}[/tex] = [tex]\frac{100*4}{2000}[/tex]
= [tex]\frac{400}{2000}[/tex]
= 0.2
[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]
The area of the small piston is 0.2 [tex]m^{2}[/tex].
Coulomb's Law says that the forces between any two charges depends ___________.
Question 7 options:
A. Directly on the size of the charges
B. Inversely on the square of the distance between the charges
C. Directly on the square of the distance between the charges
D. A&B
E. A/C
Answers
Hope this helps!!!!!!
Coulomb's Law says that the forces between any two charges depends directly on the size of the charges and inversely on the square of the distance between the charges. The correct option is D. A & B
Coulomb's lawFrom the question, we are to determine the correct option.
First, we will state Coulomb's law.
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects
That is,
[tex]F \propto \frac{Q_{1} Q_{2} }{r^{2} } [/tex]
Where F is the electrical force
[tex]Q_{1} \ and \ Q_{2}[/tex] are the charges
r is the distance between the objects
From the law, we can observe that the forces depends directly on the size of the charges - Option A
and the forces also depends inversely on the square of the distance between the charges - Option B
Hence, the correct option is D. A & B
Learn more on Coulomb's law here: https://brainly.com/question/16033085
Una pieza de platino metalico con densidad 21.5 g/cm3 tiene un volumen de 4.49 cm3. Cual es su masa
Answers
Answer:
m = 96.53 grams
Explanation:
Given that,
The density of metallic platinum, d = 21.5 g/cm³
Volume, V = 4.49 cm³
We need to find the mass. Let it is m. We know that the density is equal to the mass per unit volume. So,
[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\m=21.5\times 4.49\\\\m=96.53\ g[/tex]
So, the required mass is 96.53 grams.
What is the only part in our body we do not stretch?
Answers
Answer:
The quadriceps consists of four muscles: the skinny rectus femoris and the three huge “vasti” — vastus lateralis/intermedius/medialis. The vasti are only elongated by knee flexion, which is limited to about 120˚ when the calf hits the hamstrings. The vasti cannot be stretched strongly.
Answer:Masseter and temporalis
Why it’s unstretchable: The jaw can only open so far.
Why it’s a dang shame: Jaw tension is epidemic, and trigger points in these muscles cause a wide array of strange face and head pains, including toothaches, headaches, and earaches.
On planet Q, g = 2.24 m/s2. If the mass of planet Q is 8.96
1021 kg, what is the radius of planet Q?(G = 6.67 * 10-11 N
(m/kg))
a
8.95 x 1032 m
b 5.17 x 10 m
C 1.26 108 m
d 9.23 x 1024 m
e
4.72 x 106 m
Answers
Answer:
5.17 x 10^5
Explanation
The correct answer is (b) [tex]5.17*10^{5}m[/tex]
Let us consider an object of mass m on the planet Q.
On planet Q, let the acceleration due to gravity be g = 2.24 m/[tex]s^{2}[/tex]
Mass of the planet is M = [tex]8.96*10^{21} kg[/tex]
Let its radius be R
and the universal gravitational constant G = [tex]6.67*10^{-11} m/kg[/tex]
Gravitational Force acting on the object:
[tex]mg=\frac{GMm}{R^{2} }[/tex]
[tex]R=\sqrt{\frac{GM}{g} } \\R=\sqrt{\frac{6.67*10^{-11}*8.96*10^{21} }{2.24} }\\R=5.17*10^{5} m[/tex] is the radius of the planet Q.
Learn more about gravitational force:
https://brainly.com/question/17531377
What is the main reason people and bears slip and fall on ice, be careful when outside in the snow/ice.
Answers
Answer:
Both ice and snow have a lower coefficient of friction in comparison with grass, wood or pavement.
Explanation:
The main reason why people and bears, slip and fall on ice/snow is that both ice and snow have a lower coefficient of friction in comparison with grass, wood or pavement. We present below a comparative chart of coefficients of friction associated with different materials:
Material Coefficient of Friction - Static
Ice/Snow 0.05 - 0.3
Wood 0.2 - 0.6
Asphalt 0.5 - 0.7
Grass 0.3 - 0.5
An 6 kg object accelerating from 17 m/s to 10 m/s. What is the change in momentum of the object?
Answers
Answer:
- 42
Explanation:
m = 6kg
vi = 17 m/s
vf = 10 m/s
delta mv = ?
Formula
delta mv = m*(vf) - m(vi)
delta mv = 6(10 - 17)
delta mv = 6(-7)
delta mv = - 42 kg m/s
If the net force acting on an object is zero, its inertia is also zero.
True or false
Answers
Answer:
its false
Explanation:
can someone please help:)
1- A positive charge of 3x10-7 is located in a field of 27N/C directed toward the south. What is the force acting on the charge?
2- A positive test charge of 5x10-6Cis in an electric field that exerts a force of 2x10-4N on it.
What is the magnitude of the electric field at the location of the test charge?
Answers
Explanation:
(1) Given that,
A charge, [tex]q=3\times 10^{-7}\ C[/tex]
Electric field, E = 27 N/C
We need to find the force acting on the charge. The force on the charge is given by :
[tex]F=qE\\\\F=3\times 10^{-7}\times 27\\F=8.1\times 10^{-6}\ N[/tex]
So, the force acting on the charge is [tex]8.1\times 10^{-6}\ N[/tex]
(b) Charge, [tex]q=5\times 10^{-6}\ C[/tex]
Force, [tex]F=2\times 10^{-4}\ N[/tex]
Let E be the electric field.
[tex]E=\dfrac{F}{q}\\\\E=\dfrac{2\times 10^{-4}}{5\times 10^{-6}}\\\\E=40\ N/C[/tex]
So, the electric field is 40 N/C.