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2 years ago

What is the displacement if I walk three blocks east two blocks south one block west and two blocks north

Physics

1 answer:

vladimir1956 [14]2 years ago

4 0

Answer:

south east

Explanation:

because two step to the east then a step to the west displaces one to east still but southwards

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You are tasked with calibrating the springs for a pinball machine. Your method of testing the springs is by attaching masses ont

choli [55]

Answer:

$490.5\ \text{N/m}$

Explanation:

m = Mass attached to spring = 14 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

x = Displacement of spring = $78-50=28\ \text{cm}$

k = Spring constant

The force balance of the system is given by

$kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{14\times 9.81}{0.28}\\\Rightarrow k=490.5\ \text{N/m}$

The spring constant for that spring is $490.5\ \text{N/m}$ .

6 0

2 years ago

A step-down transformer has more loops on which coil?

Nesterboy [21]

A step-down transformer has more loops in : A. Primary coil

Primary coil refers to the coil to which alternating voltage is supplied. It's usually connected to the AC supply

hope this helps

6 0

3 years ago

Read 2 more answers

a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by

lisov135 [29]

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

$v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s$

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

4 0

2 years ago

Question 8

FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

8 0

2 years ago

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago