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svp [43]
2 years ago
14

Express in words AND mathematically the relationship between… Period and frequency

Physics
1 answer:
castortr0y [4]2 years ago
6 0
Frequency = how many waves you get per sec
Period = how long each wave takes

Period = 1/frequency
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If your friend drops a chocolate bar to you from a height of 5.0 m above your hands,
Sladkaya [172]

Answer:

<h3>1.01 s</h3>

Explanation:

Using the equation of motion S = ut+1/2gt² to solve the problem where;

u is the initial velocity of the chocolate = 0m/s

t is the time taken

g is the acceleration due to gravity = 9.81m/s²

S is the height of fall = 5.0m

Substituting the given parameter into the formula to get the time t we have;

5 = 0(t)+1/2(9.81)t²

5 = 4.905t²

t² = 5/4.905

t² = 1.019

t = √1.019

t = 1.009 secs

<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>

6 0
3 years ago
Which of the five themes of geography would a person most likely use to study the importation of Chinese goods into the United S
Morgarella [4.7K]
B. Movement would be the correct answer.
7 0
2 years ago
The weight of a rock is 8.5 pounds. Does this statement include
Kaylis [27]

Answer:

quantitative

Explanation:

4 0
2 years ago
projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor
SIZIF [17.4K]

Answer:

Range, R = MV²/2QE

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

5 0
3 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
2 years ago
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