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3 years ago

Does the size of the particles change as the substance changes state?

Physics

2 answers:

Digiron [165]3 years ago

7 0

Answer:

Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas.

Explanation:

When substances change state there is no change in mass so if 100 g of ice is melted 100g of water are formed this will boil to form 100g of steam (this is called "conservation of mass").

Lelechka [254]3 years ago

4 0

Answer:

No. The size of the particles in the substance remains the same when the state of matter changes.

Explanation:

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A circuit has two 4Ω resistors in series with each other. What is the total resistance?

Paha777 [63]

Answer: 8Ω

Explanation:

Since there are two resistors of 4Ω

connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i.e Rtotal = R1 + R2

R1 = 4Ω

R2 = 4Ω

Rtotal = ?

Rtotal = 4Ω + 4Ω

Rtotal = 8Ω

Thus, the total resistance of the circuit is 8Ω

3 0

3 years ago

The figure (Figure 1) shows the velocity of a solar-powered motorhome (RV) as a function of time. The driver accelerates from a

SpyIntel [72]

Explanation :

It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.

We know that acceleration is defined as the rate of change of velocity.

$a=\dfrac{dv}{dt}$

Where

dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s

dt is the change in time, dt = 40 s - 30 s = 10 s

So, $a=\dfrac{-60\ m/s}{10\ s}$

$a = -6\ m/s^2$

From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. $-6\ m/s^2$ .

6 0

3 years ago

A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00

Doss [256]

Answer:

d = 105 m

Explanation:

Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

d = v × t

d = 21 m/s × 5 s

d = 105 m

So, it will cover 105 m before it stops.

5 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

Read 2 more answers

What part of the electromagnetic spectrum can be seen by humans

Amanda [17]

Answer:

Visible light

Explanation:

Electromagnetic spectrum is the classification of the electromagnetic waves according to their frequency/wavelength. In order from the shortest to the longest wavelength, we have

Gamma rays

X-rays

Ultraviolet

Visible light

Infrared

Microwaves

Radio waves

All these waves are invisible to human eye, except for the part referred as 'visible light'. The electromagnetic waves of this part of the spectrum are visible to human eye, and they appear as a different color depending on their wavelength. In particular, we have:

Violet: 380-450 nm

Blue: 450-495 nm

Green: 495-570 nm

Yellow: 570-590 nm

Orange: 590-620 nm

Red: 620-750 nm

4 0

3 years ago