Which describes one feature of the image formed by a convex mirror?????

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:

see below

Explanation:

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

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

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

This question is incomplete, the complete question is;

In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background signal level is 13 mV, the signal is 91 mV with no attenuator and is 25 mV with an attenuator in place. Calculate pt /pi

Answer: the pt/pi is 15.3846%

Explanation:

Given that;

Background = 13 mV

corrected signal = 91 mV - 13 mV = 78 mV

with attenuator = 25 mV - 13 mV = 12 mV

pt/pi = ?

so pt/pi = 12/78  × 100

= 15.3846%

Therefore the pt/pi is 15.3846%

Answer:

false

Explanation:

ke sath hi hai j clin oncol biol to u h

Answer:

Increase

Explanation:

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

Answer:

C and D

Explanation:

Answer:

C. at point 2 the concentrations of H20...O2... & H2 are all changing

toward their equilibrium concentrations

D. at point 1 more reactants are converted to products than

products are converted to reactants

Answer: 1.67 x 10^-7N

Explanation:

Answer:

[tex] \frac{1}{ - 729} [/tex]

Explanation:

(-243)^-6/5

[1/(-243)] ^6/5

[tex]( \frac{1}{ - 243} ) {}^{6 \div 5} [/tex]

[tex] (\frac{1}{( - 3)} ) {}^{5} ) {}^{6 \div 5} [/tex]

[tex]( \frac{1}{ - 3} ) {}^{6} [/tex]

[tex] \frac{1}{ - 729} [/tex]

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

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 work done by the catcher is 120 J.

Explanation:

Given;

velocity of the fastball, v = 40 m/s

mass of the fastball, m = 150 g = 0.15 kg

Based on work-energy theorem, the work done by the catcher is equal to the kinetic energy of the fastball.

The kinetic energy of the fastball is given as;

K.E = ¹/₂mv²

K.E = ¹/₂ x 0.15 x 40²

K.E = 120 J

Therefore, the work done by the catcher is 120 J.

Answer:

first is regular reflection and 2nd is irregular

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:

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:

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:

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:

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:

C

Explanation:

Some of the mass

Answer:

D. Mass was transformed into energy during the process.

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:

a. Concave mirror, radius of curvature = 16 cm b. magnification = 2

Explanation:

a. Since the image is upright and larger than the object, we need a concave mirror.

We know image height, h'/object height, h = -image distance, d'/object distance, d

h'/h = -d'/d

Using the real is positive convention,

h'= + 5.4 cm, h = + 2.7 cm and d = + 12 cm.

So, + 5.4 cm/+ 2.7 cm = -d'/+ 12 cm

2 = -d'/12

d' = -2 × 12 cm

= -24 cm

Using the mirror formula 1/d + 1/d' = 2/r where r = radius of curvature of the mirror

1/+12 + 1/- 24 = 2/r

1/12(1 - 1/2) = 2/r

1/12(1/2) = 2/r

1/24 = 2/r

r/2 = 24

r = 2 × 24

r = 48 cm

b.

magnification = image height, h'/object height,h = + 5.4 cm/+ 2.7 cm = 2

Answer:

it depends on the appliance

Explanation:

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

Answer:

The time to attain the maximum angular velocity is 0.33 s.

Explanation:

Given;

angular acceleration of the fan, α = 150 rad/s²

maximum angular velocity, [tex]\omega_f[/tex] = 50 rad/s

When a fan is switched on, the initial angular velocity, [tex]\omega_i = 0[/tex]

The time to attain the maximum angular velocity is given by;

[tex]\omega_f = \omega_i +\alpha t\\\\50 = 0 + 150t\\\\t = \frac{50}{150}\\\\t = 0.33 \ s[/tex]

Therefore, the time to attain the maximum angular velocity is 0.33 s.

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.

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:

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:

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