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Ostrovityanka [42]

2 years ago

13

How to install goodman heat pump with ducting apartment

1 answer:

Helga [31]2 years ago

8 0

Answer:

Explanation:

Use a pump for that u poor man

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A 60.0 kg soccer player kicks a 0.4000 kg stationary soccer ball with 6.25 N of force. How fast does the soccer ball accelerate,

frozen [14]

F = ma
6.25 N = 0.4 kg · a
a = (6.25/0.4) m/s² since N=kg·m/s²
a = 15.625 m/s²

The answer is c) 15.6 m/s²
(Note that the mass of the soccer player is irrelevant.)

8 0

3 years ago

What did Thomson's and Rutherford's experiments have in common?

sdas [7]

<span>They both used charged particles in their experiments.</span>

4 0

3 years ago

Read 2 more answers

A 5.0 kg medicine ball and a 7.0 kg medicine ball are the same size and shape. If you were in outer space, how would you determi

Scrat [10]

Answer:
The objects can be distinguished by their weight instead of by mass.

Explanation
Because mass is constant, the two objects cannot be distinguished by mass.

However, gravitational acceleration varies in outer space. Therefore the heavier mass will register a higher reading on a weighing scale.

Note that an object of mass M weighs Mg, where g = acceleration due to gravity.

5 0

3 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

bija089 [108]

Answer:

$v_o\approx0.7059\ m.s^{-1}$

Explanation:

Given:

mass of the boat, $m=120\ kg$

uniform speed of the boat, $v=1\ m.s^{-1}$

rate of accumulation of water mass in the boat, $\dot m=100\ kg.hr^{-1}$

time of observation, $t=0.5\ hr$

The mass of the boat after the observed time:

$m_o=m+\dot m\times t$

$m_o=120+100\times 0.5$

$m_o=170\ kg$

<u>Now using the conservation of momentum:</u>

$m.v=m_o.v_o$

$120\times 1=170\times v_o$

$v_o\approx0.7059\ m.s^{-1}$

4 0

3 years ago