When you turn off your bedroom lights, the kitchen lights can stay on. This is because your home is wired using which of the following? (AKS 10b / DOK 1)

A.
Fuses

B.
Series circuits

C.
Batteries

D.
Parallel circuits

Answer:

D- Z, X, Y, W

Explanation:

Correct on edge

Answer:

[tex]\mid \vec C\mid=31.9[/tex]

Explanation:

Consider the vectors:

[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]

[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]

Calculate the magnitude of

[tex]\vec C=-2\vec B-\vec A[/tex]

Substitute the values of the vectors:

[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]

Operate and remove parentheses:

[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]

Operating both components separately:

[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]

Now find the magnitude of C:

[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]

[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]

[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]

Answer:

F =  3.32 x 10⁻⁶ N

Explanation:

The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:

F = Gm₁m₂/r²

where,

F = Force between Romeo and Juliet = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of Romeo = 68 kg

m₂ = mass of space station = 4.58 x 10⁵ kg

r = distance = 25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²

F =  3.32 x 10⁻⁶ N

Answer:

15 m/s

Explanation:

The Speed Of The Car Does Not Add To The Speed Of The Bullet

Answer:

they are cooler than the rest if the sun

The force required to hold it completely submerged under water is 0.252 N

As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.

Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation

Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m

Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³

Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg

Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N

Volume of water displaced = 4/3 π r³ = 2.805 e-5

Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N

F = 0.275 - 0.023 = 0.252 N

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The force required to hold it completely submerged under water is 0.25 N

The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³

The density of water [tex]\rho_w[/tex] = 1000 kg/m³

Diameter = 3.77 cm = 0.0377 m

radius of ball = 0.0377/2 = 0.01885 m

The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]

Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:

The force is required to hold it completely submerged under water (F) is:

[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]

F = 0.25 N

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It is the proportion predicted to be present in the early universe.

The hydrogen and helium abundance helps us to model the expansion rate of the early universe.

In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.

In big Bang model, the universe is mostly light or protons.

The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.

Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all  it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.

Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.

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Answer:

Explanation:

Frequency is oscillations per second.

So we have to find the Number of seconds she paced.

2 minutes = 2 X 60

               = 120 seconds

Therefore,

Her frequency = 10 / 120

                       = 1/12 Hertz

From a flying plane a body should dropped in advance to hit the target,Why? ... The body should be dropped in advance as when the body is dropped it has the velocity of the plane. So, in air the body moves forward which we have to take into consideration in order to hit the target.

Answer:

yeah

Explanation:

yeah yeah yeah yeah

Answer:

Th average force impact is [tex]F = 168.298 \ N[/tex]

Explanation:

From the question we are told that  

   The mass of the golf ball is  [tex]m_g = 26.7 \ g = 0.0267 \ kg[/tex]

    The angle made is [tex]\theta = 33.6 ^o[/tex]

    The range of the golf ball  is [tex]R = 190 \ m[/tex]

     The duration of contact is [tex]\Delta t = 7.13 \ ms = 7.13 *10^{-3} \ s[/tex]

Generally the range of the golf ball is mathematically represented as

       [tex]R = \frac{v^2 sin2(\theta)}{g}[/tex]

Here v  is the velocity with which the golf club propelled it with, making  v the subject

       [tex]v = \sqrt{\frac{R * g}{sin 2 (\theta)} }[/tex]

=>     [tex]v = \sqrt{\frac{190 * 9.8}{sin 2 (33.6)} }[/tex]

=>     [tex]v = 44.94 \ m/s[/tex]

Generally the change in momentum of the golf ball is mathematically represented as

      [tex]\Delta p = m * (v - u )[/tex]

here u  is the initial  velocity of the ball before being stroked and the value is 0 m/s

       [tex]\Delta p = 0.0267 * ( 44.94 - 0 )[/tex]

=>    [tex]\Delta p = 1.19996 \ kg \cdot m/s[/tex]

Generally the  average force of impact is mathematically represented as      

         [tex]F = \frac{\Delta p }{\Delta t}[/tex]

=>        [tex]F = \frac{1.19996 }{7.13 *10^{-3}}[/tex]

=>        [tex]F = 168.298 \ N[/tex]

We know, for single slit :

[tex]y =\dfrac{ n\lambda L}{a}\\\\\lambda = \dfrac{ya}{nL}[/tex]       ...1)

[tex]y = 1.4\ mm = 1.4 \times 10^{-3}\ m[/tex]

n = 2

L = 89 cm = 0.89 m

[tex]a=7.1\times 10^{-4}\ m[/tex]

Putting all these in equation 1), we get :

[tex]\lambda = \dfrac{ya}{nL}\\\\\lambda = \dfrac{1.4\times 10^{-3}\times 7.1\times 10^{-4}}{2\times 0.89 }\\\\\lambda = 5.584 \times 10^{-7}\ m[/tex]

Therefore, wavelength of the incident light is [tex]5.584 \times 10^{-7}\ m[/tex] or 558.4 nm.

Hence, this is the required solution.

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

Answer:

How long do you jump (sec) ?

Explanation:

Answer:

Fall

Explanation:

I think is True: Because...

when you throw a disc you can throw it in any direction, many people call it "Flying Saucer" I'll give you an example ... When you throw something, for example a paper, you want to throw it at your classmate. You already know what address you want to send it to, then I say it is: True ...

Sorry if it's wrong :(

Answer:

Yes

Explanation:

I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.

HOPE THIS HELPED

Answer:

Explanation:

The total work done by the wave is expressed as;

Workdone = Potential energy + Kinetic energy

Workdone = mgh + 1/2mv²

m is the mass = 77kg

g is the acceleration due to gravity = 9.8m/s²

v is the velocity = 8.2m/s

h is the height = 1.65m

Substitute into the formula;

Workdone = 77(9.8)(1.65) + 1/2(77)8.2²

Workdone = 1245.09 + 2588.74

Workdone = 3833.83Joules

Hence the amount of non conservative work done on the sofa is 3833.83Joules

Given:

We know,

→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]

or,

                     [tex]= mgh +\frac{1}{2} mv^2[/tex]

By putting the values,

                     [tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]

                     [tex]= 1245.09+2588.74[/tex]

                     [tex]= 3833.83 \ Joules[/tex]

Thus the above approach is right.

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Answer:

The total energy is 53 Joule

Explanation:

Mechanical Energy

The mechanical energy of an object of mass m, speed v, and at a height h is:

[tex]\displaystyle E = m.g.h+\frac{mv^2}{2}[/tex]

The ball has a mass of m=0.5 Kg, a speed v = 4 m/s, and at a height of h=10 m. Thus the total energy is:

[tex]\displaystyle E = 0.5\cdot 9.8\cdot 10+\frac{0.5\cdot 4^2}{2}[/tex]

E = 49 J + 4 J = 53 J

The total energy is 53 Joule

Answer:

(a) V₂ = 1.86 m/s

(b) Q = 5.1 x 10⁻⁴ m³/s

Explanation:

(a)

The formula derived for Venturi tube is as follows:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where,

P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa

ρ = Density of Gasoline = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

Therefore,

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²   ------------------- equation (1)

Now, we will use continuity equation:

A₁V₁ = A₂V₂

where,

A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

Therefore,

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

using this value in equation (1):

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

(b)

For fluid flow rate we use the following equation:

Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)

Q = 5.1 x 10⁻⁴ m³/s

The formula for finding variables in a  Venturi tube is shown below:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity

P₁ - P₂ = 1.2 KPa = 1200 Pa

ρ = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

We then substitute the variables into this equation.

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²  ------ equation (1)

The  continuity equation A₁V₁ = A₂V₂ is then used

where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

We then substitute the value into equation 1

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

This can be calculated using the formula

Flow Rate = Q = A₂V₂

                  = (2.746 x 10⁻⁴ m²)(1.86 m/s)

                 = 5.1 x 10⁻⁴ m³/s

I would say food but like they take weight too

We are given:

Homer the Astronaut:

Mass of Homer the astronaut(m1) = 200 kg

initial velocity of Homer the astronaut(u1) = 5 m/s

Larry the Pal:

Mass of Larry the Pal (m2) = 150 kg

initial velocity of Larry the Pal (u2) = 0 m/s

Since they will move together after the collision, they will have the same velocity:

v1 = v2 = V

Solving for the Final velocity:

from the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

since v1 = v2 = V:

m1u1 + m2u2 = V(m1 + m2)

replacing the variables with the given values

200 * 5 + 150 * 0 = V(200 + 150)

1000 = 350V

V  = 1000 / 350

V = 2.86 m/s

Answer:

The earliest known life-forms are putative fossilized microorganisms, found in hydrothermal vent precipitates, that may have lived as early as 4.28 Gya (billion years ago), relatively soon after the oceans formed 4.41 Gya, and not long after the formation of the Earth 4.54 Gya.

Explanation:

Is this what your looking for?

Answer:

The total thermal energy required  is 8.93 Kcal

Explanation:

Given;

mass of the ice cube, m = 43 g

specific heat capacity of water, Cp = 4.18 J/gc

specific latent heat of fusion of ice, Cf = 334 J/g

First step, determine the heat needed to raise the temperature of the ice from -16.5 °C to 0° C

Q₁ = mCp[0 - (-16.5)]

Q₁ = 43 x 4.18(16.5)

Q₁ = 2965.71 J

Second step, determine the latent heat of fusion of ice at 0°C

Q₂ = mCf

Q₂ = 43 x 334

Q₂ = 14362 J

Third step, determine the quantity of heat required to raise the temperature of the water initially at 0°C to above 11.5 °C of boiling point of water.

The final temperature of the water = 11.5 °C + 100° C = 111.5 °C

Q₃ = mCp Δθ

Q₃ = 43 x 4.18 (111.5 - 0)

Q₃ = 20041 J

Total thermal energy required = Q₁ + Q₂ + Q₃

Total thermal energy required =  2965.71 J + 14362 J +  20041 J

Total thermal energy required = 37,368.71 J

Total thermal energy required = 8.93 Kcal

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

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Answer:

a. Object A

Explanation:

The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.

The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.

So that,

For object A on the moon, mass = 25 lbs

For object B on the earth, W = 25 N,

W = m x g

25 = m x 10                (g = 10 m/[tex]s^{2}[/tex])

m = [tex]\frac{25}{10}[/tex]

   = 2.5 lbs

Mass of object B is 2.5 lbs.

Therefore, the mass of the object A is more than that of B.