Momentum is defined as mass ___ velocity

Answer:

560.54 Ω

Explanation:

Applying,

V = IR'............... Equation 1

Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors

make R' the subject of the equation

R' = V/I............ Equation 2

From the question,

Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A

Substitute these values into equation 2

R' = 5/(2.23×10⁻³ )

R' = 2242.15 Ω

Since the fours resistor are connected in series and they are equal,

Therefore the values of each resistor is

R = R'/4

R = 2242.15/4

R = 560.54 Ω

Answer:

a)[tex]|\Delta E|=4.58\: J[/tex]  

b)[tex]F=61.90\: N[/tex]

Explanation:

a)

We can use conservation of energy between these heights.

[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]  

[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]

Therefore, the lost energy is:

[tex]|\Delta E|=4.58\: J[/tex]  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

[tex]W=\Delta E[/tex]

[tex]F*d=mgh[/tex]

Let's solve it for F.

[tex]F=\frac{mgh}{d}[/tex]

[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]

Therefore, the force is:

[tex]F=61.90\: N[/tex]

I hope is helps you!

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]

[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]

Therefore, the number of turns of wire needed is 573.8 turns

Explanation:

Given that,

Amplitude, A = 2.5 nm

Wavelength,[tex]\lambda=1.8\ m[/tex]

The speed of the wave, v = 36 m/s

At time t = 0 the left end of the string has its maximum upward displacement.

(a) Let f is the frequency. So,

[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]

(b) Angular frequency of the wave,

[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]

(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]

[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.

[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

[tex]I = \frac{ E}{ R + r} \\[/tex]

[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]

Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have

[tex]➺ \: I \: (R + r) = E \times 1[/tex]

[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]

[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]

[tex]\boxed{ Hence\:proved. }[/tex]

[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]

Answer:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

Answer:

Decreases

Explanation:

because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again

Complete question is;

A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly

downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

Answer:

Power ≈ 600,000 W

Explanation:

We are given;

Frequency; f = 2400 Hz

height of the oven cavity; h = 25 cm = 0.25 m

base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²

total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J

We want to find the power output and we know that formula for power is;

P = workdone/time taken

Formula for time here is;

t = h/c

Where c is speed of light = 3 × 10^(8) m/s

Thus;

t = 0.25/(3 × 10^(8))

t = 8.333 × 10^(-10) s

Thus;

Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))

Power ≈ 600,000 W

Explanation:

15.556 metres per second

Answer:

5.04 m

Explanation:

You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.

Answer:

5.0384m

Explanation:

% increase = 100 x (Final - Initial / | initial | )

( |~~| Bars indicate absolute value since you can't have a negative height)

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Answer:

Option (e)

Explanation:

A body executing SHM moves to and fro or back and forth  about its mean position.

When the particle is at mean position, its velocity is maximum and when it is at extreme position its velocity is zero.

So, when it is at maximum distance:

a.

The acceleration is maximum.

b.

The potential energy is maximum.

c.

The total mechanical energy is non zero.

d.

The kinetic energy is zero.

e. The speed is zero. Correct

Answer:

the weight of the air in pound-force (lb-f) is 325.8 lbf

Explanation:

Given;

dimension of the room, = 15 ft by 15 ft by 20 ft

density of air in the room, ρ = 0.0724 lbm/ft³

The volume of air in the room is calculated as;

Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³

The mass of the air is calculated as;

mass = density x volume

mass = 0.0724 lbm/ft³  x  4,500 ft³

mass = 325.8 lb-m

The weight of the air is calculated as;

Weight = mass x gravity

Weight = 325.8 lb-m x 32.174 ft/s²

Weight = 10482.29 lbm.ft/s²

The weight of the air in pound-force (lb-f) is calculated as;

1 lbf = 32.174 lbm.ft/s²

[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]

Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf

Relative density is the ratio of the density of a substance to the density of a given material.

Answer:

Answer to the following question is as follows;

Explanation:

The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.

Explanation:

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

Answer:

1) attached below

2) assumption that the earth is spherical

Explanation:

1) Four illustrations of a globe

attached below

2) Reason for distortions at areas that do not fall on lines of tangency or secancy

The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true

Answer:

Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)

Answer: 11.36 m

Explanation:

Given

Mass of roller coaster is m=1200 kg

Initial speed of roller coaster is v=15 m/s

Energy at bottom and at the top is same i.e.

[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]

Thus, the highest point reach by the roller coaster is 11.36 m

Answer:

11.36m

Explanation:

Answer:

(a) Iron > plastic > ice

(b) Same on all

(c) Iron downwards, plastic net force zero, ice upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.

Explanation:

Three spheres have same volume , plastic, ice and iron.

(a) The weight is given by

Weight = mass x gravity = volume x density x gravity

As the density of iron is maximum and the density of ice is least so the order of the weight is

Weight of iron > weight of plastic > weight of ice

(b) Buoyant force is given by

Buoyant force = Volume immersed x density of fluid x g

As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.

(c) Net force is

F = weight - buoyant force  

So, the net force on the iron sphere is downwards

On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.  

Answer:

Hence the amount of heat transfer is 918.75 Btu.

Explanation:

Now,

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N

F = Kq₁q₂ / r²

Where

From Coulomb's law,

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ = constant

F₁r₁² = F₂r₂²

With the above formula, we can obtain the final force as follow:

F₁r₁² = F₂r₂²

1.2 × r² = F₂ × (2r)²

1.2r² = F₂ × 4r²

Divide both side by 4r²

F₂ = 1.2r² / 4r²

F₂ = 0.3 N

Learn more about Coulomb's law:

https://brainly.com/question/506926

Full Question:

What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?

A) 0°F

B) 273 K

C) 0 K

D) 100°C

E) 273°C

Answer:

The correction Option is D) 100°C

Explanation:

The temperature above is referred to as the critical point.

it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.

There is also a condition under which water can exist in its three forms: that is  

- Ice (solid)

- Liquid (fluid)

- Gas (vapor)

That state is called triple point. The conditions necessary for that to occur are:

Cheers

Cheers

Answer:

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Explanation:

The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:

[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)

Where [tex]t[/tex] is the time, measured in seconds.

The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:

[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)

Then we proceed to integrate on the function in time:

[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]

[tex]\Delta \theta = 20\,rad[/tex]

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Answer:

a

Explanation:

Answer:

the answer is c

'

Explanation: