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Veseljchak [2.6K]
3 years ago
13

PLEASE HELP!!

Physics
1 answer:
astraxan [27]3 years ago
3 0

Answer:

Explanation:

\lambda\\ = v/f

^That is the formula we are going to use.

Now, we were given the speed (v), which is 20.

Now we need to find frequency, in order to solve for the wavelength.

Frequency is the amount of waves in a fixed unit of one second, meaning our F value is the value of 5 divided by 4.

5/4 = 1.25

Therefore our F is 1.25

Now lets plug it in

\lambda\\ = v/f

\lambda\\ = 20/1.25

\lambda\\ = 16

Conversion:

\lambda\\ = 8

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Gold has a specific heat of 129 J/kg°C. How many joules of heat energy are required
ikadub [295]
To raise 1 kg by 1C, it will take 129 joules.

To raise 0.015 kg by (85-22)C, it will take (129*0.015)*22 = 42.57 joules.
6 0
4 years ago
Vector A has magnitude 8.00 mm and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
Ilia_Sergeevich [38]

Answer:

-75.35°

Explanation:

Let C be the sum of the two vectors A and B. Hence, we can write the following  

A_{x} +B_{x} =C_{x} ......(1)\\A_{y} +B_{y} =C_{y} ......(2)

but since the vector C is in the -y direction, C_{x}  = 0 and C_{y} = —12 m.  

Thus  

B_{x} =-A_{x} =-[-Acos(180-127)]=(8)*cos(53)\\B_{x} =4.81m

similarly, we can determine B_{y} by rearranging equation (1)  

 B_{y} =C_{y} -A_{y} =-12m-[(8)*sin(53)\\B_{y} =-18.4m

so the magnitude of B is

B=\sqrt{B_{x}^2+B_{y}^2  } \\B=19m

Finally, the direction of B can be calculated as follows  

Ф=tan^{-1} (\frac{B_{y} }{B_{x} } )\\=-75.35

hence the vector B makes an angle of 75.35 clockwise with + x axis

8 0
4 years ago
Which of the following best describes the shape of Earth's orbit?
nikklg [1K]

Elliptical, as shown by most projections


4 0
3 years ago
Describe what happens at the molecular level during melting
Strike441 [17]
The molecules of a solid vibrate faster so that they spread out to become a liquid hope this helps
5 0
4 years ago
A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
4 years ago
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