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ollegr [7]
3 years ago
15

Letícia leaves the grocery store and walks 150.0 m to the parking lot. Then, she turns 90° to the right and walks an additional

70.0 m to her car. What is the magnitude of the displacement of her car from the grocery store exit? Round your answer to the nearest tenth.
Physics
2 answers:
kvv77 [185]3 years ago
4 0
So you need to find the resulting length between the two distances. If you look at the two vectors as components of a right triangle then you can use Pythagorean theorem to solve for the last side. it is simply \sqrt{150^{2}+70^{2}}  which equals 165.5
Novosadov [1.4K]3 years ago
4 0

Answer:

165.5 m

Explanation:

In this problem, we have two displacements:

- A first displacement of 150.0 m forward

- A second displacement of 70.0 m to the right

The two displacement are in perpendicular direction, so in order to calculate the resultant displacement, we can use the Pytagorean theorem (in fact, the two displacements are the sides of a right triangle, and the hypothenuse corresponds to the magnitude of the resultant displacement).

Therefore, the magnitude of her total displacement is:

R=\sqrt{d_1 ^2 +d_2^2}=\sqrt{(150.0 m)^2+(70.0 m)^2}=165.5 m

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Answer:

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3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La
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Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

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f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}    ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

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you replace the values of L, M and T in the expression (1) for getting f1:

f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz

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4 0
3 years ago
1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i
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Answer: 2.67 m/s

Explanation:

Given

Mass of block A  is m_a=2\ kg

mass of block B is m_b=3\ kg

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the initial velocity of block B is u_b=0

After collision velocity of block A is v_a=1\ m/s

Conserving momentum

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The momentum of block A after the collision is P_a=2\times 1=2\ kg.m/s

Therefore, there is no change in sign.

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