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3 years ago

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Determine how would the frequency of the pendulum change if it was taken to the moon by finding the ratio of its frequency on th

e moon fM to its frequency on the earth fE. Suppose that gE is the free-fall acceleration on the earth and gM is the free-fall acceleration on the moon.
Express your answer in terms of some or all of the variables l, m, gE, gM.
fM/fE = ?

1 answer:

andrezito [222]3 years ago

7 0

Answer:

you lady boo

Explanation:

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oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

puteri [66]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

Heat transfer from a hot surface to a cool surface in direct contact is called

navik [9.2K]

This is an example of conduction

8 0

3 years ago

A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co

nignag [31]

Answer:

Explanation:

Given

mass of box $m=80 kg$

coefficient of kinetic friction $\mu _k=0.15$

coefficient of Static friction $\mu _s=0.30$

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

$f_r=ma$

where $f_r=frictional\ Force$

$a=acceleration$

$a=0$

$f_r=0$

4 0

3 years ago

a car is traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is the acceleration?

torisob [31]

I can’t see if anyone have gave you the answer yet

5 0

3 years ago

Read 2 more answers