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2 years ago

Un corredor hace los 400 metros lisos en 50 seg. Calcula la velocidad en la carrera.

Physics

1 answer:

RoseWind [281]2 years ago

6 0

Answer:

8 meters/second

Explanation:

Rate x Time = Distance

R x T = D

R x 50sec = 400meters

R x 50sec/50sec = 400meters/50sec

R = 8meters/sec

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Explain biome in your own words.

EleoNora [17]

Hello,

A biome is a big habitat consisting of biotic and abiotic factors. there are 7 main biomes they are tundra.

taiga.

temperate deciduous forest.

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3 0

3 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

5 0

3 years ago

A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.

Simora [160]

Answer:

$\mu_k=\frac{a}{g}$

Explanation:

The force of kinetic friction on the block is defined as:

$F_k=\mu_kN$

Where $\mu_k$ is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

$N=mg\\F_k=F=ma$

Replacing this in the first equation and solving for $\mu_k$ :

$ma=\mu_k(mg)\\\mu_k=\frac{a}{g}$

6 0

3 years ago

One runner in a relay race has a 1.50 s lead and is running at a constant speed of 3.25 m/s. The runner has 30.0 m to run before

yarga [219]

The second runner must run 3.3m/s. If the leading runner is 1.5 seconds ahead and there are 30m left, the second runner would need to run slightly faster than the lead in order to finish at the same time. To calculate this I did 30/1.5 which gave me 0.05. I added this onto the speed of the lead runner to get 3.3m/s :)

6 0

3 years ago

Read 2 more answers

Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loa

Alex787 [66]

Answer:

(a)

W_friction = 98.1 J

W_net = 550.9 J

(b)

W_friction = 98.1 J

W_net = 463.95 J

Explanation:

(a)

First, we will calculate the work done by friction:

$W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\$

W_friction = 98.1 J

Now, the work done by Eskimo will be:

$W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 0^o\\$

W_Eskimo = 649 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 649 J - 98.1 J

W_net = 550.9 J

(b)

First, we will calculate the work done by friction:

$W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\$

W_friction = 98.1 J

Now, the work done by Eskimo will be:

$W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 30^o\\$

W_Eskimo = 562.05 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 562.05 J - 98.1 J

W_net = 463.95 J

4 0

2 years ago