Which trophic levels does a lizard have on Food Web

On May 20, 1999, 37-year old Robbie Knievel, son of the famed daredevil Evel

Ostrovityanka [42]

R/RocketLeague - Every Painted Uncommon Wheels (Pics)
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Tired of trading some of the new wheels and not having a pic of them? Want to collect some sweet wheels but you don't know which? Say no more! The complete collection of the painted 17 different uncommon wheels it's here.

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6 0

3 years ago

A current of 6.93 A 6.93 A in a long, straight wire produces a magnetic field of 4.15 μT 4.15 μT at a certain distance from the

natka813 [3]

Answer:

0.334 m

Explanation:

The magnetic field due to current in a straight thin wire, $B$ , is given by

$B = \dfrac{\mu_0 I}{2\pi R}$

where $\mu_0$ is the permeability of free spave with value $4\pi\times10^{-7}$ and $R$ is the distance from the wire.

$R = \dfrac{\mu_0 I}{2\pi B}$

Substituting the values from the question,

$R = \dfrac{4\pi\times10^{-7}\times 6.93}{2\pi\times 4.15\times 10^{-6}}$

$R = \dfrac{2\times10^{-1}\times6.93}{4.15} = 0.334 \text{ m}$

7 0

4 years ago

Read 2 more answers

An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

$M=-\frac{q}{p}$

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

$M=-\frac{200 cm}{50 cm}=-4.0$

7 0

3 years ago

In order to cool 1 ton of water at 20°C in an insulated tank, a person pours 140 kg of ice at –5°C into the water. The specific

salantis [7]

Answer:

$T = 6.31 ^0 C$

Explanation:

1 ton = 907.2 kg

so here heat given by water = heat absorbed by the ice

so by energy balance we will have

$Q_1 = Q_2$

$mL + ms\Delta T_1 + ms\Delta T_2 = ms\Delta T_3$

so we have

$140(2.11)(5) + 140(333.7) + 140(4.18)(T) = 907.2(4.18)(20 - T)$

$48195 + 585.2 T = 75841.9 - 3792.1 T$

$T = \frac{27646.92}{4377.3}$

$T = 6.31 ^0 C$

7 0

3 years ago

A mass m at the end of a spring vibrates with a frequency

Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

$\omega = \sqrt{\frac{k}{m}}$

And given that ω = 2πf

the relation between frequency and mass becomes

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$

6 0

3 years ago