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maks197457 [2]
2 years ago
14

Select all that apply.

Physics
2 answers:
Bad White [126]2 years ago
6 0
Star s has radial motion
rjkz [21]2 years ago
3 0

Answer:

Star S has radial motion

Explanation:

Because the radial motion stracture is kinda similar to hydrogen spectrum(I'm guessing,only lol)

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Can someone help me with this question
monitta

Answer:

hypothesis , hope it helps

Explanation:

7 0
4 years ago
Read 2 more answers
A box weighing 43.2 N is pulled horizontally until it slides uniformly lat a constant
GREYUIT [131]

Your diagram should include four forces:

• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)

• the normal force, pointing up (mag. <em>n</em>)

• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")

• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )

The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have

<em>n</em> + (-<em>w</em>) = 0

and

<em>p</em> + (-<em>f</em> ) = 0

So then the forces have magnitudes

<em>w</em> = 43.2 N

<em>n</em> = <em>w</em> = 43.2 N

<em>p</em> = 6.30 N

<em>f</em> = <em>p</em> = 6.30 N

5 0
3 years ago
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

4 0
3 years ago
How is the density of a substance calculated?
den301095 [7]
1).  Take a sample of the substance.  The sample should be the largest
possible that will allow it to be be easily handled and the following steps
to be performed with it. 
(The density doesn't depend on the size of the sample, and every sample
of the same substance has the same density.  But using a larger sample
can improve the accuracy of the measurements you make, and therefore
improve the accuracy of the density you derive for the substance.)

2). Ask or measure the mass of the sample.

3). Ask or measure the volume of the sample.

4). Divide the mass by the volume. Their quotient is the density
of the substance.
3 0
3 years ago
A bicyclist travels $4.5\text{ km}$ west, then travels $6.7\text{ km}$ at an angle $27.0^\circ$ South of West.
MatroZZZ [7]

Answer:

10.88 km

Explanation:

We shall represent displacement in terms of i , j  unit vectors in the direction of east and north .

4.5 km due west

D₁ = - 4.5 i

6.7 km at an angle of 27° south of west

D₂ = - 6.7 cos27 i - 6.7 sin27j

= - 6.7 x .89 i - 6.7 x .45 j

= - 5.96i - 3 j

Total displacement

= D₁ + D₂

=  - 4.5 i - 5.96i - 3 j

= -10.46 i - 3j

Magnitude = √ ( 10.46² + 3²)

= √ ( 109.41 + 9)

= √ 118.41

= 10.88 km .

7 0
3 years ago
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