A pump lifts 400 kg of water per hour a height of 4.5 m .
Part A
What is the minimum necessary power output rating of the water pump in watts?
Express your answer using two significant figures.
Part B
What is the minimum necessary power output rating of the water pump in horsepower?
Express your answer using two significant figures.

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]

So, the approximate pressure is equal to 40.69 atm.

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

Answer:

[tex]K.E_{max}=0.8973J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.64kg[/tex]

Equation of Mass

[tex]X=0.33cos(3.17t)[/tex]...1

Generally equation for distance X is

[tex]X=Acos(\omega t)[/tex]...2

Therefore comparing equation

Angular Velocity [tex]\omega=3.17rad/s[/tex]

Amplitude A=0.33

Generally the equation for Max speed is mathematically given by

[tex]V_{max}=A\omega[/tex]

[tex]V_{max}=0.33*3.17[/tex]

[tex]V_{max}=1.0461m/s[/tex]

Therefore

[tex]K.E_{max}=0.5mv^2[/tex]

[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]

[tex]K.E_{max}=0.8973J[/tex]

Answer:

idek

Explanation:

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

Explanation:

Given:

L = 0.02 H

C = [tex]2\:\mu \text{F}[/tex]

f = 200 Hz

The general form of the impedance Z is given by

[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to

[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]

[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

Answer:

By the principle of corona discharge.

Explanation:

The charge on each ball will decreases over time due to the electrical discharge in air.

According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.

Answer:

1. CaO is not an organic compound because it doesn’t contain a carbon molecule.

2.

Name General Structure Properties/Uses

Alcohol R-OH (contains a hydroxyl group)  Can be poisonous, can be made from fermentation or distillation

Aldehyde R-COH (contains a carbon atom double-bonded to an oxygen and single-bonded to a hydrogen)  Makes up formaldehyde and acetaldehyde

Ketone R-CO-R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds)   Makes up acetone

Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters

Ether R-O-R (contains double carbon chains connected to an oxygen atom through single bonds)   Ethyl ether is very volatile and flammable, used in veterinary medicine

3. Carbon is able to make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.

Explanation:

pf

CaO is not an organic compound because it doesn’t contain a carbon molecule.

(which contains a hydroxyl group)  Can be poisonous, can be made from fermentation or distillation

Aldehyde R-COH (contains a carbon atom double-bonded to oxygen and single-bonded to hydrogen)  Makes up formaldehyde and acetaldehyde

Ketone R-CO-1R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds)   Makes up acetone

Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters11

Ether -O-R (contains double carbon chains connected to an oxygen atom through single bonds)   Ethyl ether is very volatile and flammable, used in veterinary medicine

Carbon can make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.

Learn more about organic molecules.

https://brainly.com/question/24225576

#SPJ2

Answer:

The correct answer will bs "2.41 m".

Explanation:

According to the question,

M = 50 g

or,

   = 0.050 kg

[tex]\Theta = 25^{\circ}[/tex]

k = 25.9 N/m

Δx = 0.200 m

Let the traveled distance be "x".

By using trigonometry, the height will be:

⇒ [tex]h = l Sin \Theta[/tex]

hence,

⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]

                                       [tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]

By putting the values, we get

             [tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]

                                              [tex]l=2.41 \ m[/tex]      

Answer:

have an increased resistance

Answer:

[tex]X=0.0389m[/tex]

Explanation:

From the question we are told that:

Period of spring [tex]T_s=2.25s[/tex]

Initial Position of Mass [tex]x=0.0480m[/tex]

Final Mass period [tex]T_f=5.85s[/tex]

Generally the equation for the Mass location is mathematically given by

[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]

[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]

[tex]X=0.0389m[/tex]

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]

Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]

Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]

This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.

Answer:

The intensity of 4 lawn movers is 86 dB.

Explanation:

Intensity of one lawnmower = 80 dB

Let the intensity is I.

Use the formula of intensity

[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]

Now the intensity of 4 lawn movers is

[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]

If an icy surface means no friction, then Newton's second law tells us the net forces on either block are

• m = 1 kg:

∑ F (parallel) = mg sin(45°) - T = ma … … … [1]

∑ F (perpendicular) = n - mg cos(45°) = 0

Notice that we're taking down-the-slope to be positive direction parallel to the surface.

• m = 0.4 kg:

∑ F (vertical) = T - mg = ma … … … [2]

Adding equations [1] and [2] eliminates T, so that

((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a

(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a

==>   a ≈ 2.15 m/s²

The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.

Answer:

219 ft

Explanation:

Here we can define the value t = 0s as the moment when the car starts decelerating.

At this point, the acceleration of the car is given by the equation:

A(t) = -10 ft/s^2

Where the negative sign is because the car is decelerating.

To get the velocity equation of the car, we integrate over time, to get:

V(t) = (-10 ft/s^2)*t + V0

Where V0 is the initial velocity of the car, we know that this is 88 ft/s

Then the velocity equation is:

V(t) = (-10 ft/s^2)*t + 88ft/s

To get the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0

Where P0 is the initial position of the car, we do not know this, but it does not matter for now.

We want to find the total distance that the car traveled in a 3 seconds interval.

This will be equal to the difference in the position at t = 3s and the position at t = 0s

distance = P(3s) - P(0s)

 = ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)

=  ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)

=  (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft

The car advanced a distance of 219 ft in the 3 seconds interval.

Answer:

7.8% of the original volume.

Explanation:

From the given information:

Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C

Pressure [tex]P_1[/tex] = 240 kPa

Temperature [tex]T_2[/tex] = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

[tex]P_1V_1 =nRT_1[/tex]

making V_1 (initial volume) the subject:

[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]

[tex]V_1 = \dfrac{nR*295}{240}[/tex]

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume [tex]V_2[/tex] can be computed as:

[tex]V_2 = \dfrac{nR*318}{240}[/tex]

Now, the change in the volume ΔV =  V₂ - V₁

[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]

[tex]\Delta V = \dfrac{23nR}{240}[/tex]

∴

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]

[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]

[tex]= 0.078[/tex]

= 7.8% of the original volume.

Answer:

true

Explanation:

im not sure please dont attack me

Answer:

  v = 2.75 10⁴ m / s

Explanation:

For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system

            F = ma

where force is the force of gravity

            F = [tex]G \frac{m M}{r^2}[/tex]

acceleration is centripetal

             a = [tex]\frac{v^2}{r}[/tex]

we substitute

           G m M / r² = m v² / r

           [tex]\frac{GM}{r}[/tex] = v²

           v = [tex]\sqrt{GM/r}[/tex]

indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system

            r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m

let's calculate

         v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11

         v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8

         v = 2.75 10⁴ m / s

Answer:

Test 1

1.True

2.True

3.True

4.False

5.True

6.True

7.False

8.True

9.True

10.True

yung iba nasa pic

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].

Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].

Solids cannot diffuse.

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N

Answer:

d

Explanation:

Ya gon find the Kenitic Energy first

K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004

and now the replacement:

0.004=½×0.4V²====> v²=0.02===> V=0.14m/s

Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]

By rearrangement and making (v) the subject of the above formula:

[tex]v = \dfrac{uf}{u-f}[/tex]

replacing the given values:

[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]

[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]

v = 0.857 cm

Answer:

pretty sure its 6.2 x 10^-13

Explanation:

I looked it up I'm not a bigbrain but want to help