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3 years ago

Maggie completed a 10000-m race at an average speed of 160

Physics

1 answer:

Gala2k [10]3 years ago

3 0

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

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Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe

NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B ) Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

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3 years ago

Read 2 more answers

8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer

Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

$M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \ by \ the \ effort}{disatnce \ moved \ by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m$

Therefore, the distance that the object is raised above its initial position is 5.625 m.

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2 years ago

It takes 5.0 seconds for a wave with a wavelength of 2.5 m to travel past your location. What’s the period of the wave. What is

Anna35 [415]

(a) Period of the wave
The period of a wave is the time needed for a complete cycle of the wave to pass through a certain point.
So, if an entire cycle of the wave passes through the given location in 5.0 seconds, this means that the period is equal to 5.0 s: T=5.0 s.

(b) Frequency of the wave
The frequency of a wave is defined as
$f= \frac{1}{T}$
since in our problem the period is $T=5.0 s$ , the frequency is
$f= \frac{1}{5.0 s}=0.2 Hz$

(c) Speed of the wave
The speed of a wave is given by the following relationship between frequency f and wavelength $\lambda$ :
$v=f \lambda = (0.2 Hz)(2.5 m)=0.5 m/s$

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