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3 years ago

5

Adhesion is when forces of the molecules are more strongly drawn to each other than the molecules in other substances true or fa

lse

1 answer:

SSSSS [86.1K]3 years ago

6 0

Answer:

False

Explanation:

that is cohesion. adhesion is force between dissimilar molecules of a body

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Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of oneâs birth. The on

Anton [14]

Answer:

Explanation:

Gravitational force between two objects having mass m₁ and m₂ at a distance R

F = G m₁ m₂ / R²

Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²

= 1.01 x 10⁻⁶ N

b )

Force between baby and Jupiter

F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²

= 1.31 x 10⁻⁶ N

c )

Ratio = 1.01 / 1.31

= .77

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3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

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3 years ago

At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta

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2 years ago

A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute

Amiraneli [1.4K]

it can be said that the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

what is the speed of the east wind?.

<h3> the speed of the east wind</h3>

Generally the equation for the distance is mathematically given as

BA=3000sin60

BA=2598.07m

Therefore

the speed of the east wind

$V_w=\frac{BA}{120*60}\\\\V_w=\frac{2598.07}{120*60}$

v=0.3608

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2 years ago

Please somebody answer I desperately need help!

padilas [110]

Preserved fossil<span> (like a fossil in amber, ice or tar.</span>

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3 years ago

Read 2 more answers