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3 years ago

Help would be appreciated.

Physics

1 answer:

Kaylis [27]3 years ago

7 0

Answer:

10m

Explanation:

1000kg/m³÷100KPa

=10m

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What is the acceleration of the object?<br> m/s²

Orlov [11]

Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5

7 0

2 years ago

Two molecules of lithium are combined with 225 grams of bromine to form two molecules of lithium bromide. If you end up with 690

4vir4ik [10]

The amount of Li present to start the reaction is 55.18g

<u>Explanation:</u>

2Li + Br₂ → 2LiBr

Molecular weight of Br₂ = 159.808 g/mol

Mass of Br₂ present = 225 g

Moles of Br₂ present during the reaction = 225 / 159.808

m = 1.4

Molecular weight of LiBr = 86.845 g/mol

Mass of LiBr formed = 690 g

moles of LiBr produced = 690 / 86.845

m(LiBr) = 7.95

According to the balanced equation, 2 molecules of Li reacts to for 2 molecule of LiBr

So, 7.95 moles of LiBr would require 7.95 moles of Li

The molecular weight of Li is 6.941 g/mol

Thus, the amount of lithium present to start the reaction is

$moles = \frac{given weight}{molecular weight} \\\\7.95 = \frac{w}{6.941} \\\\w = 55.18g$

Therefore, the amount of Li present to start the reaction is 55.18g

6 0

3 years ago

My model is supposed to have

bearhunter [10]

Answer:

yes you can submit

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4 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

Two people are pushing a couch from opposite sides. one person pushing to the right with a force of 3 N. the other person pushes

skelet666 [1.2K]

letter a because 4N to the left +3N to the right = 4N left minus 3N left as you change the direction of 3N then you also change the plus into minus. the direction will always favour to which of the direction is bigger

8 0

3 years ago