3 years ago

What does aerobic refer to?

Physics

2 answers:

Gnesinka [82]3 years ago

7 0

Answer:

A) how your body uses oxygen

svetoff [14.1K]3 years ago

4 0

It’s the first option

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Two iron bolts of equal Mass one at a hundred see another at 55 Sierra place in the insulated cylinder assuming the heat capacit

malfutka [58]

Answer:

$T_2 = 77.5c$

Explanation:

From the question we are told that

Temp of first bolts $T_1=100$

Temp of 2nd bolt $T_2=55$

Generally the equation showing the relationship between heat & temperature is given by

$q=cm \triangle T$

Generally heat released by the iron bolt = heat gained by the iron bolt

Generally solving mathematically

$-(0.45*m* (T_2-100 \textdegree c)) = 0.45*m*(T_2 -55\textdegree c)$

$-(T_2-100 \textdegree c)) = (T_2 -55 \textdegree c)$

$T_2 +T_2= 100 \textdegree c+55 \textdegree c$

$T_2=\frac{155 \textdegree c}{2}$

$T_2 = 77.5 \textdegree c$

Therefore $T_2 = 77.5 \textdegree c$ is the final temperature inside the container

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2 years ago

If the energy from the light is captured and stored by the surface without being reflected or transmitted, then it has been ____

vekshin1

A. ABSORBEDDDDDDDDDDDDDDDDDD

6 0

3 years ago

Read 2 more answers

Select the correct answer. Which of Newton's laws explains why your hands get red when you press them hard against a wall? A. Ne

STatiana [176]

Answer:

D newton third law

Explanation:

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2 years ago

20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?

Tresset [83]

Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance
so Work done = 0.085 x 20g x 28
= 466.956 J

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3 years ago

Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the

Natasha_Volkova [10]

The force per unit length between the two wires is $6.0\cdot 10^{-5} N/m$

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0 = 4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

$I_1, I_2$ are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

$I_1 = 1.75 A$

$I_2 = 3.45 A$

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

$\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m$

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3 years ago