What is matter ?


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Answer:

Seek shelter under stable tables or under door frames.

If outside, stay away from buildings, bridges and electricity pylons and move to open areas.

Avoid areas at risk from secondary processes, such as landslides, rockfall and soil liquefaction.

From  

Measures against earthquakeswww.planat.ch › earthquake › measures-eb

Answer: 5.91kw

Explanation:

given data:

power = 65kw

voltage stepped up = 120v t0 1200

R 0.100

Solution:

when the it is stepped to 120v

[tex]P = VI \\65000/ 120I\\ I = 541.7A\\[/tex]

power loss

[tex](I^{2} )R*2\\= (541.7^{2})0.100*2\\= 58536.2w[/tex]

when it is stepped to 1200v

[tex]\frac{65000}{1200} \\ I = 54.17A\[/tex]

[tex](I^{2} )R*2\\= (54.17^{2})0.100*2\\= 586.9w[/tex]

total power saved

[tex]= 65kw - 58.5kw - 0.59kw\\= 5.91kw[/tex]

Answer:

Explanation:

.... ...

Answer:

see below

Explanation:

 It boils (or evaporates)  and becomes gaseous nitrogen like water turns to water vapor when it gets warm enough

Answer:

I don't know if its just me, but I don't see the statements so I'm not sure how to help answer your question.

Answer:

Image B

Explanation:

although I'm not exactly sure, i've recently gotten this question as well. but model B demonstrates the force- distance trade off because you can see how in that image them distance is increased in the force is decreased with the object being shorter. hopefully this helps in some way

Answer:

C

Explanation:

GPS satellites in Earth's orbit constitute the most common way that images of space are captured.

I hope this helps! And this is the real answer.

Answer:

C

Explanation:

Some of the mass

Answer:

D. Mass was transformed into energy during the process.

Answer:

The maximum height reached by the projectile is 490.31 km.

Explanation:

Given;

initial velocity of the projectile, u = 3.1 km/s = 3100 m/s

acceleration due to gravity, g = 9.8 m/s²

The maximum height reached by the projectile is calculated as;

v² = u² - 2gh

where;

v is the final velocity of the projectile at maximum height = 0

h is the maximum height reached by the projectile

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (3100²) / (2 x 9.8)

h = 490306.122 m

h = 490.31 km

Therefore, the maximum height reached by the projectile is 490.31 km.

Answer:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

[tex]\Delta\ U = nC_{v}\Delta\ T[/tex]

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

[tex]\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J[/tex]

Now, we can apply First Law of Thermodynamics as follows:

[tex]\Delta\ Q = \Delta\ U + W[/tex]

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

[tex]-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\[/tex]

W = - 118.24 J (negative sign shows that work is done on piston)

Answer:

-49.2

Explanation:

Trust me bro

Answer:

Explanation:

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer

Using the formula for calculating speed expressed as;

Speed = Distance/Time

Given;

Distance = 3km = 3000m

Speed = 1000m/s

Required

How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer (Time)

From the formula;

Time = Distance/speed

Time = 3000/1000

Time = 3seconds

Hence the sound of the aircraft is heard after 3 seconds

Answer:

Increase

Explanation:

If an object is falling for a greater distance, it will gain momentum and speed up.

Answer:

false

Explanation:

ke sath hi hai j clin oncol biol to u h

Answer:

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

Given that,

The equation is "F=kx"

To find,

The derivative equation.

Solution,

We are given with the given equation which is as follows :

F = kx ...(1)

Where, F is force, k is spring constant and x is distance covered by the spring

Differentiate equation (1) wrt x.

[tex]\dfrac{dF}{dx}=\dfrac{d}{dx}(kx)\\\\\dfrac{dF}{dx}=k[/tex]

As k is constant.

Hence, this is the required solution.

Answer:

The acceleration acting on the train during this

time of travel is 0.85m/s²

HOW TO CALCULATE ACCELERATION:

• The acceleration of a moving body can be calculated by using the formula below:

a = (v - u)/t

Where;

1. a = acceleration (m/s²)

2. v = final velocity (m/s) 3. u = initial velocity (m/s)

4. t = time (s)

According to this question, a = ?, v = 48m/s, u = 14m/s, t = 40s.

= (48 - 14)/40 a=

• a = 34/40

• a = 0.85m/s²

Therefore, the acceleration acting on the train during this time of travel is 0.85m/s².

Hope this helps!

Don't forget to mark me as Brainliest.

Answer:

I=0.02[amp]

Explanation:

Since the resistors are connected in series, we can find the total resistance.

[tex]Rt = 100 + 300+50\\Rt = 450[ohm][/tex]

Now we can use ohm's law which tells us that the voltage is equal to the product of resistance by the current.

[tex]V =I*R\\I=V/R\\I=9/450\\I=0.02[amp][/tex]

Answer:

it depends on the appliance

Explanation:

bigger appliances will use more power, smaller appliances will use a lesser amount

Answer:

A Living things can only come from living things.

B Cells come from pre-existing cells.

Explanation:

sorry if wrong e d g e 2022

Answer: D,E

Explanation:

I JUST TOOK IT

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = [tex]\pi r^2[/tex]

V = Potential applied = 2 mV

k = Dielectric constant = 40

[tex]\epsilon_0[/tex] = Electric constant = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]

Capacitance is given by

[tex]C=\dfrac{k\epsilon_0A}{d}[/tex]

Charge is given by

[tex]Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}[/tex]

Number of electron is given by

[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}[/tex]

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

Explanation:

Objects at rest and in motion respond to the presence of an external unbalanced force by simple changing their magnitude of motion or position.

We have this knowledge from Newtons first law of motion "a body will remain in a state of rest or continue with uniform motion unless if it is acted upon by an external force".

Answer:

10 Pa

Explanation:

Pressure is defined as force per unit area.

Mathematically, P= F/A where  P is pressure, F is force and A is area.

Force, F = 60 N

Area, A = 2*3 = 6m²

P = 60/6 = 10 Pa

Answer: 1.67 x 10^-7N

Explanation:

Answer:

first is regular reflection and 2nd is irregular

a = v² / R = (2 m/s)² / (0.1 m) = 40 m/s²

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

[tex] P_{1} = P_{2} [/tex]

[tex] m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

[tex]v_{1_{i}}[/tex]: is the initial velocity of the ball 1 = 6.20 m/s

[tex]v_{2_{i}}[/tex]: is the initial velocity of the ball 2 = 0 (it is at rest)

[tex]v_{1_{f}}[/tex]: is the final velocity of the ball 1 =?

[tex]v_{2_{f}}[/tex]: is the initial velocity of the ball 2 =?

[tex] m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]

[tex] v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}} [/tex] (1)        

Now, by conservation of kinetic energy (since they collide elastically):

[tex] \frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2} [/tex]          

[tex] m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2} [/tex]  (2)

By entering equation (1) into (2) we have:

[tex] m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2} [/tex]    

[tex] 0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2} [/tex]            

By solving the above equation for [tex]v_{2_{f}}[/tex]:

[tex]v_{2_{f}} = 3.1 m/s [/tex]

Now, [tex]v_{1_{f}}[/tex] can be calculated with equation (1):

[tex] v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s [/tex]

The minus sign of [tex] v_{1_{f}}[/tex] means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!                  

Answer:

Approximately [tex]1.69 \times 10^{-23}\; {\rm N}[/tex].

Explanation:

Look up the value of Coulomb's Constant: [tex]k \approx 8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}}[/tex].

Consider point charges of magnitude [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. If the distance between these charges is [tex]r[/tex], the magnitude of the electrostatic force between them would be [tex](k\, q_{1}\, q_{2}) / (r^{2})[/tex].

In this question, the two [tex](+q)[/tex] charges are [tex]5\; {\rm cm}[/tex] and [tex]3\; {\rm cm}[/tex] away from the center [tex](-2\, q)[/tex] charge, respectively. Convert units to standard unit of distance (meters, [tex]{\rm m}[/tex]) and charge (coulombs, [tex]{\rm C}[/tex]):

[tex]q = 1.15 \; {\rm nC} = 1.15 \times 10^{-9}\; {\rm C}[/tex].

[tex]\begin{aligned} 5\; {\rm cm} = 5\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.05\; {\rm m} \end{aligned}[/tex].

[tex]\begin{aligned} 3\; {\rm cm} = 3\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.03\; {\rm m} \end{aligned}[/tex].

The magnitude of the electrostatic forces on the [tex](-2\, q)[/tex] charge would be:

[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.05\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 9.509\times 10^{-24}\; {\rm N}\end{aligned}[/tex].

[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.03\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 2.641\times 10^{-23}\; {\rm N}\end{aligned}[/tex].

Since the charges are of opposite sign, the [tex](-2\, q)[/tex] charge would attract both of the [tex](+q)[/tex] charges. In particular, the (approximately) [tex]9.509\times 10^{-24}\; {\rm N}[/tex] force would point to the left. The (approximately) [tex]2.641 \times 10^{-23}\; {\rm N}[/tex] force would point to the right.

As a result, the net force on the [tex](-2\, q)[/tex] charge would point to the right. The magnitude of the net force on this charge would be approximately [tex]2.641 \times 10^{-23}\; {\rm N} - 9.509\times 10^{-24}\; {\rm N} \approx 1.69 \times 10^{-23}\; {\rm N}[/tex].