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3 years ago

I need to know which ones are physical or chemical changes.

Physics

1 answer:

andrezito [222]3 years ago

7 0

Physical
Chemical
Physical
Chemical
Chemical
Chemical
Physical
Physical
Physical
Chemical
Chemical
Physical
Physical
Physical
Chemical
Chemical
Chemical
Chemical
Physical
Chemical
Chemical
Chemical
Chemical
Chemical
Chemical

I believe this is right but double check to make sure :)

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Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,

jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

Therefore: $E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}$

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$

7 0

3 years ago

A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'

NNADVOKAT [17]

Answer:

Acceleration, $a=-1.5\ m/s^2$

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{-5-10}{10}$

$a=-1.5\ m/s^2$

So, the acceleration of the car is $-1.5\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the grou

spayn [35]

Answer:

154° at 195 km/h

Explanation:

The helicopter is moving south at 175 km/h, relative to the wind.

But the wind is moving east at 85 km/h, relative to the ground.

This means that the helicopter is moving south east relative to the ground.

Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.

This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.

Refer to the triangle b1.

The distance traveled by the helicopter in 1 hour is denoted by d.

d is the hypotenuse of the right triangle.

Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)

Hence the helicopter is traveling at 195 km/h relative to the ground.

To calculate the direction we use,

tan (x) = opposite/adjacent = 85/175

So the angle x is,

$x = arctan (\frac{85}{175} )$ = 25.9°

Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)

8 0

3 years ago

Explain whether the unit of work is a fundamental or derived unit

klasskru [66]

Answer:

answer here

Explanation:

the unit of work is fundamental unit because it doesn't depend on other units.

__________________

Thx

6 0

3 years ago

Read 2 more answers

The y component of the electric field of an electromagnetic wave travelling in the +x direction through vacuum obeys the equatio

notsponge [240]

Answer:

λ =8.57 μ m

Explanation:

Given that

Ey = 375 cos [kx − (2.20 × 10¹⁴ rad/s)t] N/C

Standard form

Ey=Eo cos[k x-ωt] N/C

By comparing the given equation with the standard wave equation

Eo = 375 N/C

ω = 2.20 × 10¹⁴ rad/s

We know that ω = 2 π f

$f=\dfrac{\omega}{2\pi }$

$f=\dfrac{2.2\times 10^{14}}{2\pi }\ Hz$

f=3.50×10¹³ Hz

We know that the velocity given as

V = f λ

λ =Wavelength

V=Speed = 3 x 10⁸ m/s

$\lambda =\dfrac{V}{f }$

$\lambda =\dfrac{3\times 10^8}{3.5\times 10^{13}}\ m$

λ =0.00000857 m ( 1 μ m = 10⁶ m)

λ =8.57 μ m

8 0

3 years ago