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alukav5142 [94]
1 year ago
14

A wolf spider runs 75 cm west, then turns and runs 50 cm south. Which choice gives the correct solution for the resultant? Quest

ion 22 options: R2 = 752 + 502 R2 = 752 + 502 - 2(75)(50) cos 60 R2 = 752 502 R2 = 752 502 - 2(75)(50) cos 90
Physics
1 answer:
Serga [27]1 year ago
4 0

The correct solution for the resultant of the wolf spider is:

R² = 75² + 50²

<h3>Resultant formula</h3>

R² = (Displacement 1)² + (Displacement 2)²

Where

  • R is the resultant displcaement

See attach photo for diagram

<h3>Data obatined from the question</h3>

The following data were obtained from the question.

  • Displacement 1 = 75 cm west
  • Displacement 2 = 50 cm south
  • Resultant dispalcement (R) =?

<h3>How to determine the solution for the resultant </h3>

The solution for the resultant displacement of the wolf spider can be obtained as follow:

R² = (Displacement 1)² + (Displacement 2)²

R² = 75² + 50²

Thus, the correct answer to the question is:

R² = 75² + 50²

Learn more about resultant displacement:

brainly.com/question/13947057

#SPJ1

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Answer:

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Explanation:

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All isotopes of a particular element have the same atomic number. How then do the isotopes of a particular element differ?
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Which method will correctly determine whether the forces on an object are balanced or unbalanced?
MaRussiya [10]

For this case, the first thing you should do is define a reference system.

Once the system is defined, we must follow the following steps:

1) Do the sum of forces in a horizontal direction

2) Do the sum of forces in vertical direction

The forces will be balanced if for each direction the net force is equal to zero.

The forces will be unbalanced if for each direction the net force is nonzero.

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6 0
3 years ago
Read 2 more answers
W=90 kg×4n/kg<br>Please halp me i need it ​
kiruha [24]

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5 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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