To ____________________ something is to slowly mix in a bit of _____________ ___________________ with eggs or cream to raise the temperature slowly. This prevents the cream from curdling (turning lumpy and gross) or the eggs from scrambling in your sauce.

Answer:

6.3 C

Explanation:

From the question,

Energy lost by the capacitor(half it stored energy) = 1/4CV².......................... Equation 1

Where C = Capacitance of the capacitor, V = Volatge across the parallel plate of the capacitor.

E = 1/4CV²

Given: C = 2.1 F, V = 6.0 V

Substitute these values into equation 1

E = 1/4(2.1)(6²)

E = 18.9 J

But,

E = 1/2QV..................... Equation 2

Where Q = amount of charge lost

Make Q the subject of the equation

Q = 2E/V................. Equation 3

Give: E = 18.9 J, V = 6.0 V

Q = 2(18.9)/6

Q = 6.3 C.

Answer:

Remains the same

Explanation:

Answer:

4th answer

Explanation:

Answer: By a factor of 1/3.

Explanation:

For a photon with wavelength λ, the energy is written as:

E = h*c/λ

where:

h is the Planck's constant:

h = 6.63*10^(-34) Js

c is the speed of light:

c = 3*10^8 m/s

Now, if we triple the wavelength of this photon, then the energy will be:

E' = (h*c)/(3*λ)

We rewrite this as:

E' = (1/3)*(h*c/λ)

And (h*c/λ) was the previous energy:

(h*c/λ) = E

Then we can replace that in the above equation to get:

E' = (1/3)*(h*c/λ) = (1/3)*E

Then if we triple the wavelength, it will change the energy content of the individually radiated photons by a factor of 1/3.

Answer:

[tex]v=5m/s[/tex]

Direction:SW(south-west)

Explanation:

From the question we are told that

Velocity and direction of Runner [tex]V_r= 5m/s ,East[/tex]

Velocity and direction of Air[tex]V_a= 4m/s ,South[/tex]

Generally the the resultant velocity v is mathematically given as

[tex]v=\sqrt{V_a^2+V_r^2}[/tex]

[tex]v=\sqrt{4^2+5^2}[/tex]

[tex]v=5m/s[/tex]

The resultant velocity is towards the south-west

Answer:

1264 Hz

Explanation:

We are given;

Your distance to the front of first speaker: L1 = 2.79 m

Distance between speakers; d = 1.26m

Now, your distance to second speaker can be calculated using pythagoras theorem.

Thus;

L2 = √(1.26² + 2.79²)

L2 = 3.0613 m

Let's find the wavelength from the formula;

(L1 - L2) = nλ

Where n is the order and could be; n = 0,1,2,3...

Making λ which is the wavelength the subject, we have;

(L1 - L2)/n = λ

The least frequency will occur at the wavelength when n = 1.

λ = (2.79 - 3.0613)/1

λ = -0.2713 m

We will take the absolute value and thus;

λ = 0.2713 m

Frequency is gotten from;

f = v/λ

Where v is speed of sound = 343 m/s

Thus;

f = 343/0.2713

f ≈ 1264 Hz

Answer:

a) 75000Joules

b) 0Joules

Explanation:

Workdone = Force * Distance

Given

distance= 750m

Force = 300N

a) If the frictional force = 200N

The Total force = 300N - 200N = 100N

Work done = 100 * 750

Workdone = 75,000Joules

Hence the workdone if the force of friction is 200N is 75,000Joules

b) If the frictional force = 300N

The Total force = 300N - 300N = 0N

Work done = 0* 750

Workdone = 0Joules

Hence the workdone if the force of friction is 300N is 0Joules i.e no work will be done on the sled

Given :

The acceleration due to gravity on or near the surface of Earth is 32 ft/s/s

To Find :

From what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft/s.

Solution :

Initial velocity of stone, u = 0 ft/s.

Now, by equation of motion :

[tex]2as = v^2 -u^2 \\\\2\times 32 \times s = 136^2 -0^2\\\\s = \dfrac{136^2}{2\times 32}\ ft\\\\s = 289 \ ft[/tex]

Therefore, height from which stone is thrown is 289 ft.

Answer:

Explanation:

The bullet will hit the coconut .

Both coconut and bullet will fall under the gravitational pull of the earth . They will experience same acceleration because acceleration does not depend upon mass . So both bullet and coconut will fall by the same distance vertically .

So bullet will exactly hit the coconut .

Answer:

P = 240 W

Explanation:

       [tex]P = \frac{\Delta E}{\Delta t} (1)[/tex]

       [tex]P = \frac{\Delta E}{\Delta t} =\frac{2,400J}{10s} = 240 W (2)[/tex]

Answer:

the power used by the person is 10 W.

Explanation:

Given;

weight lifted, W = 50 N

distance through which the weight was lifted, d = 2 m

time of motion, t = 10 s

The power used by the person is calculated as;

[tex]P= FV\\\\P = F \ \times \ \frac{d}{t} \\\\P = 50 \ \times \ \frac{2}{10} \\\\P = 10 \ W[/tex]

Therefore, the power used by the person is 10 W.

Answer:

The charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg

Explanation:

From the formulae

F = qvB and F = mv²/r

Where F is Force

q is charge

v is speed

B is magnetic field strength

m is mass

and r is radius

Then,

qvB = mv²/r

qB = mv/r

We can write that

q/m = v/rB ---- (1)

Also

From Electric force formula

F = Eq

Where E is the electric field

and magnetic force formula

F = Bqv

Since, electric force = magnetic force

Then, Eq = Bqv

E = Bv

∴ v = E/B

Substitute v = E/B into equation (1)

q/m = (E/B)/rB

∴ q/m = E/rB²

(NOTE: q/m is the charge to mass ratio)

From the question,

E =  3.10 ×10³ N/C

r = 4.20 cm = 0.0420 m

B = 0.360 T

Hence,

q/m = 3.10 ×10³ / 0.0420 × (0.360)²

q/m = 569517.9306 C/kg

q/m = 5.7 × 10⁵ C/kg

Hence, the charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg.

Answer:

go to the link quizzlet it will give you tha answer

Explanation:

Answer:The powder of Kool Aid crystals are the solute. The water is the solvent and the delicious Kool Aid is the solution.... .-.

Explanation:

Answer:

22%

Explanation:

Given parameters:

The amount of change  = $0.22

Unknown

What percentage of a dollar the change represents  = ?

Solution:

A dollar is made up of 100cents;

        So;

 $0.22 represents 22cents

Percentage of dollar this represents  = [tex]\frac{22}{100}[/tex]  x 100  = 22%

Answer:

[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]

Explanation:

Given that:

[tex]\rho_o = (10^{-3} ) \ c/m^3[/tex]

R = (0.1) m

To find  the electric field for r < R by using Gauss Law

[tex]{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)[/tex]

For r < R

[tex]Q_{enclosed}=(\rho) ( \pi r^2 ) l[/tex]

[tex]E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}[/tex]

[tex]E= \dfrac{\rho ( r)}{2 \varepsilon_o}[/tex]

where;

[tex]\varepsilon_o = 8.85 \times 10^{-12}[/tex]

[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]

[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]

[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]

Answer:

F=78 N

Explanation:

      [tex]F_{app} = F_{kfr} (1)[/tex]

      [tex]F_{1kfr} = \mu_{k} * F_{n} (2)[/tex]

     where μk = coefficient of kinetic friction = 0.12

     Fn = normal force

      [tex]F_{n} = F_{g} = m*g = 650 N (3)[/tex]

      [tex]F_{1kfr} = \mu_{k} * F_{n} = 0.12 * 650 N = 78 N (4)[/tex]

       [tex]F_{app} = F_{kfr} = 78 N (5)[/tex]

Answer: A. frequency

Explanation:

If it’s not wavelength then it has to be frequency

Answer:

F = 3.2 x 10⁻⁵ N

Explanation:

The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:

[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]

where,

F = gravitational force = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m₁ = mass of object 1 = 235 kg

m₂ = mass of object 2 = 1.37 x 10¹² kg

r = distance between objects = 2.59 x 10⁴ m

Therefore,

[tex]F = \frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(235\ kg)(1.37\ x\ 10^{12}\ kg)}{(2.59\ x\ 10^{4}\ m)^{2}}[/tex]

F = 3.2 x 10⁻⁵ N

Answer:

the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Explanation:

Given;

mass of the first object, m₁ = 235 kg

mass of the second object, m₂ = 1.37 x 10¹² kg

distance between the two object, r = 2.59 x 10⁴ m

The gravitational force between the two object is calculated as;

[tex]F= \frac{Gm_1m_2}{r^2}[/tex]

where;

G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²

[tex]F= \frac{(6.67\times 10^{-11})(235)(1.37\times 10^{12})}{(2.59\times 10^4)^2} \\\\F = 3.2 \times 10^{-5} \ N[/tex]

Therefore, the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Answer:

0 N/C

Explanation:

Since the electric field due toeach charge q is at the same distance from the center of the square and has a magnitude E = 2 N/C, we resolve the components of each electric field to vertical and horizontal.

So, for each charge, the magnitude of the horizontal component is E' = Ecos45 and the magnitude of the vertical component is E" = Esin45.

For the charges on the bottom half of the square, the direction of the vertical component is upwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the bottom two charges is 2Esin45.

For the charges at the upper half of the square, the direction of the vertical component is downwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the top two charges is -2Esin45.

So, the resultant electric field at the center of the square is thus 2Esin45 + (-2Esin45) = 2Esin45 - 2Esin45 = 0 N/C  

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, [tex]v_{max}[/tex] = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

[tex]a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2[/tex]

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

[tex]a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max} \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max} \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s[/tex]

Therefore, the maximum speed is 21.39 m/s.

Answer:

65.59 m /s

Explanation:

Initial velocity u = 4.73 m/s

Final velocity v = ?

time t = 6.21 s

acceleration = g = 9.8 m /s

v = u + gt

= 4.73 + 9.8 x 6.21

= 65.59 m /s .

Answer:

"m" and "n" are 5.97 and 24 respectively.

Explanation:

Standard form which is a scientific notation [m × 10^n] can be regarded as way to reduce large figures to small one in a decimal firm and this is usually done for conviniency sake.

✓Let us find "m" , to do this we will shorten the long number so that it will be from 1-9. And this is 5.97, hence

m = 5.97

✓ " n" can be determined by counting the digits from our right hand then stop where the decimal point was put when we were finding our "m". Hence

n= 24

✓ if we input the values to the scientific expresion above, we have

m = 5.97 and n = 24

Hence, the Standard form = 5.97 × 10^24 kg which is the scientific notation.

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

Answer:

A woman with a spike heel

Explanation:

Since we're assuming here, we proceed to assuming that the mass of the elephant is 6000 kg, and the mass of the lady is 60 kg. Then,

a woman with a spike heel will exert more pressure on the ground than an elephant weighing 6000 kg. This is because, the force exerted by the woman is a 588 N and she has a, say 1 cm² spike heel. This puts half her weight on each foot, and if evenly distributed to half on her heel and half on her sole, the pressure being exerted by each heel is 147 N/ 2cm² = 147 N/cm². On the other hand, if a 58,860 N elephant with an 1000 cm² feet is exerting 1/4 its weight on each foo. Then it has 14715 N/1000 cm² = 14.7 N/cm²; which is exactly 10 times lesser pressure than what the woman is exerting. So, the woman with the hell exerts more pressure, than the elephant will.

Answer:

10.8 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration (a) = 5 m/s/s

Distance travelled (s) = 291 m

Time (t) taken =?

We can calculate the time taken for the car to cover the distance as follow:

s = ut + ½at²

291 = 0 × t + ½ × 5 × t²

291 = 0 + 2.5 × t²

291 = 2.5 × t²

Divide both side by 2.5

t² = 291 / 2.5

t² = 116.4

Take the square root of both side

t = √116.4

t = 10.8 s

Thus, it will take the car 10.8 s to cover the distance.

The time taken by the car to travel 291 m is calculated using the kinematics equation and it is obtained to be 10.8 seconds.

Here, it is given that;

The acceleration of the car is,

[tex]a = 5\,m/s^2[/tex]

The distance covered by the car is,

[tex]s = 291\,m[/tex]

Also, the car starts from rest, that means;

[tex]u = 0\,m/s[/tex]

We can use the second kinematics equation to find the time taken by the car to travel the given distance.

[tex]s = ut + \frac{1}{2}\,at^2\\291\,m = (0\,m/s \times t)+ \frac{1}{2} (5\,m/s^2\times t^2)\\\\\\291\,m = (2.5\,m/s^2\times t^2)[/tex]

implies

[tex]t = \sqrt{\frac{291\,m}{2.5\,m/s^2}}=10.78\,s\approx10.8\,s[/tex]

Learn more about 'motion in a straight line' here: https://brainly.com/question/19558242

Given :

A wooden block is let go from a height of 5.80 m.

To Find :

The velocity of the block just before it hits the ground.

Solution :

We know, by equation of motion :

[tex]v^2 - u^2 = 2as[/tex]

Here, a = g = 9.8 m/s²( Acceleration due to gravity )

Putting all given values in above equation, we get :

[tex]v^2 - u^2 = 2as\\\\v^2 -0 = 2\times 9.8 \times 5.8 \\\\v = \sqrt{2\times 9.8 \times 5.8 } \ m/s\\\\v = 10.66\ m/s[/tex]

Hence, this is the required solution.

Answer:

[tex]Q_{net}=4.38\,\,10^{-8}\,C[/tex]

Explanation:

We use Gauss's Law for the flux over a closed surface equal the net charge inside the closed surface divided the permitivity of space [tex]\epsilon_0=8.85\,\,10^{-12}\,\,\frac{C^2}{N*m^2}[/tex]

Therefore, by knowing the flux, we can estimate the net charge inside the cubic box with the product:

[tex]\Phi=\frac{Q_{net}}{\epsilon_0} \\Q_{net}=4950\,*\,8.85\,\,10^{-12} \,\,C\\Q_{net}=4.38\,\,10^{-8}\,C[/tex]

Answer: the time interval Green had to reached the top speed is 47.2 secs

Explanation:

Given the data in the question;

first we determine Green's acceleration.

from third equation of motion, v² - u² = 2as

a = v² - u² / 2s

we substitute,

a = ( (341 m/s)² - (0)² ) / 2(8047)

a = (116281 - 0) / 16094

a = 7.23 m/s²

Now we determine the time interval need to reach top or maximum speed,

v = u + at

at = v - u

t = (v - u) / a

so we substitute

t = (341 - 0) / 7.23

t = 341 / 7.23

t = 47.2 secs

Therefore, the time interval Green had to reached the top speed is 47.2 secs