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2 years ago

PLEASE TRY TO ANSWER AS MANY QUESTIONS AS YOU CAN !

Physics

1 answer:

4vir4ik [10]2 years ago

4 0

Answer:

0.81N

Explanation:

Ff = Fwx = .165x 9.81 SIn 30 = 0.81N

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How many electrons have been removed from a positively charged electroscope if it has a net charge of 6x10-11?

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Answer:

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2 years ago

Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label

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The gravitational force between two objects is given by
$F=G \frac{m_1 m_2 }{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

In this problem, $m_1 = 175 kg$ , $m_2 =15 kg$ and $r=2 m$ , therefore the gravitational force between the two objects is
$F=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(175 kg)(15 kg)}{(2m)^2}=4.38 \cdot10^{-8} N$

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2 years ago

When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point

vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

$v = Velocity \rightarrow 20m/s$

$\lambda = Wavelength \rightarrow 35*10^{-2}m$

Therefore the frequency would be given as

$f = \frac{20}{35*10^{-2}}$

$f = 57.14Hz$

The frequency is directly proportional to the angular velocity therefore

$\omega = 2\pi f$

$\omega = 2\pi *57.14$

$\omega = 359.03rad/s$

Now the maximum speed from the simple harmonic movement is given by

$V_{max} = A\omega$

Where

A = Amplitude

Then replacing,

$V_{max} = (1*10^{-2})(359.03)$

$V_{max} = 3.59m/s$

Therefore the maximum speed of a point on the string is 3.59m/s

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3 years ago

Which measure of an earthquake depends on how close you are to the focus?

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3 years ago

The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth

ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have