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3 years ago

A dog exerts a force of 30N to move a wagon 2m in 5s. What is the power of the dog

Physics

1 answer:

Hunter-Best [27]3 years ago

6 0

Explanation:

power=f×v. recall= distances/ time

= f× d/t

= 30 × 2/5

=12watt

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An atom is chemically stable when it’s outer_____ is completely filled with _____.

Svetlanka [38]

an atom is chemically stable when its outer layer is completely filled with energy. im pretty sure this is right

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3 years ago

A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus

Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

$Y=\frac{F\times L}{A\times\Delta L}$

$\Delta L=\frac{F\times L}{A\times\Y}$

$\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}$

ΔL = 1 m

Thus, the cable stretches by 1 m.

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3 years ago

Which is true of oxidation? PLS HELP FAST and thank you!

elena-14-01-66 [18.8K]

Answer:The Answer Is D

Explanation:

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2 years ago

Because the pressure falls, water boils at a lower temperature with increasing altitude. Consequently, cake mixes and boiled egg

AlexFokin [52]

Answer:

1) The boiling point of water reduces by 3.28°C at 1,000 m above sea-level

2) The boiling point of water reduces by 6.56°C at 2,000 m above sea-level

Explanation:

The variation of the boiling point of water with elevation is given as follows

The boiling point reduces by 0.5°C for every 152.4 meter increase in elevation

At sea-level, the boiling point temperature of water = 100°C

1) At 1,000 m elevation, the boiling point temperature, T = 100 - (1,000/152.4) × 0.5 ≈ 96.72 °C

Therefore, the boiling point of water reduces by 100° - 96.72° = 3.28°C at 1,000 m above sea-level

2) At 2,000 m elevation, the boiling point temperature, T = 100 - (2,000/152.4) × 0.5 ≈ 93.44°C

The boiling point of water reduces by 100° - 93.44° = 6.56°C at 2,000 m above sea-level

7 0

3 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

3 years ago