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3 years ago

Water is more dense in its solid state.

Physics

2 answers:

vagabundo [1.1K]3 years ago

8 0

Water is more dense in its solid state.

This statement is true.

il63 [147K]3 years ago

5 0

Answer: what is it asking

Explanation:

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A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

zvonat [6]

Answer:

$D_l=d$

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

$V_{e1}=V_{e2}$

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

$D=\frac{at^2}{2}$

Where

Acceleration is given as

$a=\frac{V_o}{2d}$

And

Time

$T=\frac{d}{v_0}$

Therefore horizontal displacement towards the left is

$D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}$

$D_l=d$

5 0

3 years ago

Solve the equation x=3logy2 for y.

melisa1 [442]

X = 3 · log(Y²)

X = 3 · 2·log(Y)

X/6 = log(Y)

10^(X/6) = 10^log(Y)

Y = 10^(X/6)

6 0

3 years ago

Read 2 more answers

Which order shows decreasing momentum?

klio [65]

Answer:

Explanation: Decreasing in velocity

4 0

3 years ago

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

Bad White [126]

A) Agreed.
b) Value agreed but units should be W (watts). 

c) Here's one method... 

15 miles = 24140 m 

1 gallon of gasoline contains 1.4×10⁸ J. 

So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J 

Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J 

Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J 

So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. 

Efficiency = useful work/total energy supplied 
= 1000/5800 
= 0.17 (=17%)

8 0

3 years ago

James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s

liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{James}=\sqrt{\frac{2h}{g}}$

Vertical motion of John : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{John}=\sqrt{\frac{2h}{g}}$

So both times are same.

Both of them reach the lake at the same time.

3 0

3 years ago