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3 years ago

Understands how formulas are used

Physics

2 answers:

Anna11 [10]3 years ago

5 0

Answer:

do it ur self i d k

Explanation:

Natali [406]3 years ago

4 0

Answer:

a map tell u sum to do

Explanation:

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A water wave traves with a length of 10 meters and a frequency of 30 hertz, what is the velocity of the wave?

UNO [17]

Answer: 300 m/s

Explanation:

Given that:

length of water wave (λ) = 10 metres

Frequency of water wave (F) = 30 Hertz

Velocity of water wave (V) = ?

The velocity of a wave is the distance covered by the wave in one second. It is measured in metre per second, and represented by the symbol V. It is related to frequency and wave length by the formula: V = F λ

V = 30 Hertz x 10 metres

V = 300 m/s

Thus, the velocity of the water wave is 300 m/s

4 0

3 years ago

what is the acceleration produced by the resultant force acting on an object if the coefficient of friction acting between the b

LenKa [72]

The acceleration of the block is $-0.98 m/s^2$

Explanation:

The expression for the force of friction acting on the block is (assuming the surface is horizontal and flat):

$F_f = -\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the block

g is the acceleration of gravity

and where the negative sign means the direction of the force is opposite to that of the motion of the block

If this is the only force acting on the object, then this is also the resultant force, so we can rewrite Newton's second law as:

$F=ma\\-\mu mg = ma$

where

a is the acceleration of the block

Re-arranging the equation,

$a=-\mu g$

And so this is the expression for the acceleration of the block acted upon the force of friction.

In this problem, we have:

$\mu=0.1$

$g=9.8 m/s^2$

Solving,

$a=-(0.1)(9.8)=-0.98 m/s^2$

Learn more about friction:

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#LearnwithBrainly

5 0

3 years ago

The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Fi

Leto [7]

Answer:

114.26

Explanation:

a)Formula for per unit impedance for change of base is

Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)

Zpu2: New per unit impedance

Zpu1: given per unit impedance

kV1: give base voltage

kV2: New bas votlage

kVA1: given bas power

kVA2: new base power

In the question

Zpu2=??

Zpu1= 0.3

kV2=24kV

kV1= 13.8 kV

kVA2= 1MVA ×1000= 1000 kVA

kVA1=500kVA

Zpu2= 0.3(13.8/24)²×(1000/500)

Zpu2= 0.198

b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,

Zbase= kV²/MVA

Zbase= 13.8²/(500/1000)

Zbase=380.88

Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:

Zpu=Zactual/Zbase

0.3= Zactual/380.88

Zactual= 114.26 ohms

8 0

4 years ago

A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,

Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0

3 years ago

What happens when an electron moving from the 3rd energy level to the 1st energy level?

Sholpan [36]

Answer:

A photon of wavelength 103 nm is released

Explanation:

When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.

For example, if we are talking about a hydrogen atom, the energy of the levels are:

$E_1 = -13.6 eV\\E_2 = -3.4 eV\\E_3 = -1.5 eV$

So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is

$\Delta E = E_3 - E_1 = -1.5 -(-13.6)=12.1 eV$

In Joules,

$\Delta E =12.1\cdot 1.6\cdot 10^{-19} = 1.94\cdot 10^{-18} J$

We can also find the wavelength of this photon, using the equation:

$\Delta E = \frac{hc}{\lambda}\rightarrow \lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.94\cdot 10^{-18}}=1.03\cdot 10^{-7} m = 103 nm$

7 0

4 years ago

Understands how formulas are used​

Understands how formulas are used