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3 years ago

What units are used in the periodic table for atomic mass?

Physics

2 answers:

kaheart [24]3 years ago

4 0

Atomic mass units or amu

omeli [17]3 years ago

4 0

It is usually called unified atomic mass units.

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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

8 0

3 years ago

Why do different substances have diffrent properties

Artist 52 [7]

Different elements require different levels of energy to make or break a bond

8 0

3 years ago

Read 2 more answers

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?

vampirchik [111]

Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
I = 4 A

8 0

3 years ago

Read 2 more answers

13. You push with 56 N on a 15-kg box, and there is a 23-N force of friction. How fast will the box accelerate?

Flura [38]

Answer:

The acceleration is 2.2 m/s^2

Explanation:

In the attached image, we can see the free body diagram. And using the second law of Newton it will be possible to find the acceleration of the box.

4 0

3 years ago

If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?

leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

[density * volume]_salt water = [density * volume]_fresh water ---> 1

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

h = 0.72 m (ANSWER)

8 0

4 years ago