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3 years ago

Convert 41.3 kilocalories into joules.

Physics

1 answer:

mihalych1998 [28]3 years ago

6 0

Answer:172799.2

Explanation:

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A comparison of a machine’s work output with its work input is called __??__.

Aloiza [94]

Ohhhhh its called a input machine

7 0

3 years ago

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. S

dedylja [7]

The concept that we need here to give a proper solution is mutual inductance.

The mutual inductance is given by the expression

$M=\frac{N\Phi}{I}$

Where,

I = current

N = Number of turns

$\Phi =$ Flux through the solenoid.

Part A) Then we have in our values that,

$I=6.6A$

$\Phi= 3.50*10^{-2}Wb$

$N=450$

Replacing in the equation,

$M = \frac{450*350*10^{-2}}{6.60}$

$M = 2.39H$

Part B) Here is required the Flux, then using the same expression we have that

$\Phi = \frac{IM'}{N}$

We conserve the same value for the Inductance but now we have a current of 2.6, then

$\Phi = \frac{2.6*2.39}{690}$

$\Phi = 9*10^{-3}Wb$

Therefore the flux in Solenoid 1 is $9*10^{-3}Wb$

8 0

3 years ago

A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5

koban [17]

Answer:

V = 38.0 L

Explanation:

As we know that number of moles will remains conserved inside the balloon

so we will have

$moles = \frac{PV}{RT}$

here we have

$\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}$

now we have

$P_1 = 760 mm Hg$

$P_2 = 76 mm Hg$

$V_1 = 5.00 L$

$T_1 = 20^o C = 293 K$

$T_2 = -50^o C = 223 K$

$\frac{(760mm Hg)(5L)}{R(293)} = \frac{(76mm Hg)(V)}{R(223)}$

$V = 38.0 L$

5 0

3 years ago

Read 2 more answers

1. How much work is done by a person if he lifts a load of 500 kg upto a height of 2m?

iren [92.7K]

Explanation:

if we take g=9.8m/s^2

F=mg

m=500

F=500×9.8=4900N

h=2

so w=mgh

w=4900×2

=9800j

3 0

3 years ago

MAJOR POINTS! Solve the system of equations to find <img src="https://tex.z-dn.net/?f=%20a_%7B2%7D%20%20" id="TexFormula1" title

Monica [59]

You are dealing with pulleys?

can be done with addition of the two equations to eliminate T.

$(2^{*}) \quad {} -m_1 g \cos \theta + T = m_1 a _2$
+
$(4^{*}) \quad 3m_1 g - T = 3m_1 a_2$
=

$(-m_1 g \cos\theta + 3m_1 g) + (T - T) = m_1 a _2 + 3m_1 a_2 \implies \\ \\
(-m_1 g \cos\theta + 3m_1 g) = 4m_1 a_2 \implies \\ \\
m_1(- g \cos\theta + 3g)= 4m_1 a_2$

we can cancel m₁ by dividing both sides by it, assuming mass is not zero

$(- g \cos\theta + 3g)= 4a_2 \implies \\ \\
a_2 = \dfrac{- g \cos\theta + 3g}{4} \\ \\
a_2 = \dfrac{- 9.80 \cos 60 + 3(9.80)}{4} \\ \\
a_2 = 6.125 \text{m/s}^2
$

a₂ = 6.125 m/s² ( do significant digits if you need to)

6 0

3 years ago