cylindrical in shape with a radius of 2.5 cm and a height of 6 cm. The bottle is filled with squirts of pepper extract from a machine. Each squirt fills the bottle to a height of 0.025 cm. The density of the pepper extract is 1.135 gm/mL a. The machine malfunctions and stops when the bottle is two thirds full. How many squirts had the machine squirted into the bottle at this time? Show your work. b. What is the mass of extract is in the bottle when it is two thirds full? What is the mass of 1 squirt of extract?

Answer:

mathematical model

Explanation:

A computer model is a very complex type of mathematical model.

These complex mathematical models involves series of equations or algorithms known as set of instructions for the computer to perform difficult computations. These mathematical models include dynamical systems, statistical models, or differential equations.

Answer:7.5. 10^4 J

Explanation:

Answer:

The Kinetic Energy of the car at that instant will be 123,210 Joules.

Explanation:

Given:

Mass=500 Kg

[tex]Speed= 80 km/h=\frac{80*1000}{3600}\ m/s =22.2 m/s[/tex]

Required:

Kinetic energy (K.E)=?

Formula:

[tex]K.E=\frac{1}{2}mv^{2}[/tex]

Where:

m is the mass.

v is the speed.

Solution:

[tex]K.E=\frac{1}{2}mv^{2}\\K.E=\frac{1}{2} *500* (22.2)^{2} \\K.E=123,210 J[/tex]

The Kinetic Energy of the car at that instant will be 123,210 Joules.

Answer:

Variance is 25

Explanation:

Recall that the standard deviation is defined as the square root of the variance. therefore, if you know the standard deviation [tex](\sigma)[/tex], square it and you get the variance:

[tex]Variance= \sigma^2\\Variance= 5^2= 25[/tex]

Answer:

The position at t=11 seconds is -96 m or 96 m to the left of the origin

Explanation:

Constant speed and constant acceleration motion

If a body travels the same distance per unit of time, the body has a constant speed. The equation for the distance x is:

[tex]x=v.t[/tex]

Where v is the constant speed and t is the time.

At time t=0 the body is moving at a constant speed of v=24 m/s in the positive x-direction. It keeps that condition for t=1 sec. The distance traveled is:

[tex]x=24m/s\cdot 1\ s=24\ m[/tex]

Now the body is given a constant acceleration of 6 m/s^2 in the negative x-direction, that is, a=-6 m/s^2

The distance traveled by a body at constant acceleration is given by:

[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]

Where vo is the initial speed and t is the time.

We need to find the position at t=11 sec, but we must subtract the one second already passed, thus t=10 s

[tex]\displaystyle x=24m/s\cdot 10s+\frac{-6m/s^2.(10)^2}{2}[/tex]

[tex]\displaystyle x=240\ m-360\ m=-120\ m[/tex]

This distance must be added to the actual position of the body, that is:

-120 m + 24 m = -96 m

The position at t=11 seconds is -96 m or 96 m to the left of the origin

Answer:

Based on the scientific notation ,

4.372896 x10^-7 is the answer.

Answer:

GPE = mgh = 5.00 kg .

 2.20 + 0.30 m = 122.5 J

Explanation:

Answer:

58.8 μC

Explanation:

The two balls create a dipole of dipole moment , p = qd where q = charge and d = distance between the charges = 28.9 cm = 0.289 m. The potential energy change ΔU = -W where W is the work done by the electric field of magnitude E = 592 N/C.

Now U = -pdEcosФ = -qdEcosФ where Ф is the angle between p and E. Since Ф is initially  58.9° and then becomes 0° when p and E align, the potential energy change is thus.

ΔU = -qdEcos0 - (-qdEcos58.9°) = qdE(cos58.9° - 1) = qdE(0.5165 - 1)  = -0.4835qdE.

W = - ΔU = -(-0.4835qdE) = 0.4835qdE

This work equals the kinetic energy of the systems. Since K.E = 4.86 mJ = 4.86 × 10⁻³ J.

So K.E = W

K.E = 0.4835qdE

making q subject of the formula

q = K.E/0.4835dE

q = 4.86 × 10⁻³ J/(0.4835 × 0.289 m × 592 N/C)

q = 4.86 × 10⁻³ J/82.72

q = 0.0588 × 10⁻³ C

q = 58.8 × 10⁻⁶ C

q = 58.8 μC

Answer and Explanation:

A. We have temperature t = 32

Speed of sound, s = 1087.5

As t increases by 1⁰f speed increases by 1.2

So that

S = 1088.6

T= 33⁰f

We have 2 equations

1087.5 = k(32) + c

1088.6 = k(33) + c

Subtracting both equations

(33-32)k = 1088.6-1087.5

K = 1.1

b.). S = kT + c

1087.5 = 32(1.1) + c

Such that

C = 1052.3

Therefore

S = 1.1(t) + 1052.3

C.). S = 1.1t + 1052.3

We make t subject of the formula

T = s/1.1 - 1052.3/1.1

T = 0.90(s) - 956.3

D. This means that We have temperature to rise by 0.90 whenever speed is increased

Answer:

Mass and distance

Explanation:

Answer:

Mass and distance

Explanation:

the larger the mass the faster it falls or more gravity is applied same goes for the distance the farther it is from the ground its heavier however if its to far gravity then does not apply

hope this helps :)

can have brainlist please

Answer:

1.76 m/s^2

Explanation:

The formula to find the force is F = ma (reference from 2nd law of Newton's Laws of Motion). Since you need to find the acceleration of the box, you need to use the formula a = f/m

a = 9.7 n divided by 5.5 kg = 1.76 m/s^2

Answer:

Explanation:

Voltage = 12 (the two batteries are in series and the voltage adds).

Resistance = 3 ohms

I = V / R

I = 12/3

I = 4 amperes.

None of the answers are the right ones. If there is a fifth one that reads 4 amperes then it is the correct answer.

Answer:

Diffusion is a result of the kinetic properties of particles of matter. The particles will mix until they are evenly distributed. Diffusion may also be thought of as the movement of particles down a concentration gradient. The term "diffusion" comes from the Latin word diffundere, which means "to spread out."

Explanation:

Answer:

true, there are certain forms of bacteria in water that can cause permanent brain damage (although its rare) so further contamination of water can lead to higher chances of contracting serious illnesses caused by such bacteria.

Answer:

The largest allowable Io = 5.863 * 10^26  photons

Explanation:

thickness of slab = 2.8 mm

First we express the energy of each photon

[tex]E = \frac{hc}{wavelength}[/tex]

wavelength = 313 nm

attached below is the detailed solution on how i arrived at the answer above

note : we use 2 in the integral because the slab is exposed to light from two different direction

Answer:

it would affect the present day because you changed what happened so what is happening in present time would be different. think about it this way, our lives right now and what is happening right now is all based on what has happened in the past. If something else were to happen in the past our lives right now would be different.

Answer:

it would affect the present day because you changed what happened so what is happening in present time would be different.

Explanation:

Answer:

through direct contact

Explanation:

took the test!

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

Answer:

The resistivity of this wire is approximately [tex]2.1\times 10^{-7}\; \rm \Omega \cdot m[/tex].

The conductivity of this wire is approximately [tex]4.7\times 10^6\; \rm (\Omega \cdot m)^{-1}[/tex].

Assumption: the perpendicular cross-section of this wire is circular.

Explanation:

Convert all unit to standard ones:

Assume that a wire has a resistivity of [tex]\rho[/tex], a length of [tex]l[/tex], and a cross-section area of [tex]A[/tex]. The formula for the resistance of this wire would be:

[tex]R = \displaystyle \frac{\rho \cdot l}{A}[/tex].

Rearrange this equation to find an expression for [tex]\rho[/tex], the resistivity of this wire:

[tex]\displaystyle \rho = \frac{R\cdot A}{l}[/tex].

For the wire in this question, both length [tex]l[/tex] and resistance [tex]R[/tex] are already given. However, the cross-section area [tex]A[/tex] of this wire needs to be calculated from its diameter, [tex]d[/tex]:

[tex]A = \displaystyle \pi\, \left(\frac{d}{2}\right)^2 \approx 7.08\times 10^{-8}\; \rm m^2[/tex].

Calculate the resistivity of this wire:

[tex]\displaystyle \rho = \frac{R\cdot A}{l} \approx \frac{3.0\; \Omega\times 7.08\times 10^{-8}\; \rm m^2}{1.00\; \rm m} \approx 2.1\times 10^{-7}\; \rm \Omega\cdot m[/tex].

The conductivity of a material is the reciprocal of its resistivity:

[tex]\displaystyle C = \frac{1}{\rho} \approx \frac{1}{2.1\times 10^{-7}\; \Omega\cdot m}\approx 4.7\times 10^{6}\; \rm (\Omega \cdot m)^{-1}[/tex].

Answer:

velocity

Explanation:

velocity = displacement/ time

Answer:

C₁ = 34.8 μF

C₂ = 0.2 μF

Explanation:

When the capacitors are connected in series their resultant capacitance is given by the formula:

Cs = 1/C₁ + 1/C₂

where,

Cs = Series equivalent capacitance = 4.8 μF

C₁ = Capacitance of 1st Capacitor

C₂ = Capacitance of 2nd Capacitor

Therefore,

4.8 μF = 1/C₁ + 1/C₂

(4.8 μF)(C₁ C₂) = C₁ + C₂   --------------- equation 1

When the capacitors are connected in parallel their resultant capacitance is given by the formula:

Cp = C₁ + C₂

where,

Cp = Parallel equivalent capacitance = 35 μF

Therefore,

35 μF = C₁ + C₂   -------------- equation 2

solving equation 1 and equation 2 simultaneously, we get:

C₁ = 34.8 μF

C₂ = 0.2 μF

Car B travels a distance x after time t (measured in seconds) of

x = (80 m/s) t

so that at 6:40 AM (which is 40 min = 2400 s after the starting time of 6:00 AM), it will have moved

(80 m/s) * (2400 s) = 192,000 m

If we take t = 0 to refer to 6:40 AM, the cars' respective positions x at time t are given by

A: x = (280 m/s) t

B: x = 192,000 m + (80 m/s) t

(i) After 1 h = 3600 s, their positions are

A: (280 m/s) * (3600 s) = 1,008,000 m

B: 192,000 m + (80 m/s) * (3600 s) = 480,000 m

away from the point X, so the distance between the two cars is

1,008,000 m - 480,000 m = 528,000 m = 582 km

(ii) Car A moves faster than car B, so A overtakes B as soon as their positions are equal, which happens for

(280 m/s) t = 192,000 m + (80 m/s) t

(200 m/s) t = 192,000 m

t = (192,000 m) / (200 m/s)

t = 960 s = 16 min

So after 16 min, both cars have a position of

(280 m/s) * (960 s) = 268,800 m = 268.8 km

beyond point X.

With an acceleration of 4.0 m/s², the bottle attains a speed of

(4.0 m/s²) * (9.0 s) = 36 m/s

so that the slide has length ∆x such that

(36 m/s)² - 0² = 2 * (4.0 m/s²) * ∆x

==>  ∆x = 162 m ≈ 160 m

Alternatively, we know the bottle covers a distance ∆x with acceleration a at time t according to

∆x = 1/2 a t²

so that

∆x = 1/2 * (4.0 m/s²) * (9.0 s)² = 162 m ≈ 160 m

Answer:

The value is [tex]\epsilon = 2.25 [/tex]

Explanation:

From the question we are told that

The charge acquired on each plate is [tex]q = 200 \mu C = 200 *10^{-6} \ C[/tex]

The additional charge accumulated on each plate is [tex]q_a = 250 \mu C = 250 *10^{-6} \ C[/tex]

Generally the capacitance of the capacitor before a dielectric slab is inserted is

[tex]C_i = \frac{q}{V}[/tex]

=> [tex]C_i = \frac{200 \mu C}{V}[/tex]

Generally the total charge on each plate of the capacitor when the dielectric slab was inserted is

[tex]Q = q + q_a[/tex]

=> [tex]Q = ( 200 + 250 ) \mu C[/tex]

=> [tex]Q = 450 \mu C[/tex]

Generally the capacitance of the capacitor after a dielectric slab is inserted is mathematically represented as

[tex]C = \frac{450 \mu C}{V}[/tex]

Generally the dielectric constant is mathematically represented as

[tex]\epsilon = \frac{C}{C_i}[/tex]

=>      [tex]\epsilon =  \frac{450 \mu C}{200  \mu C}[/tex]

=>       [tex]\epsilon =  2.25 [/tex]

The dielectric constant of the dielectric slab is 2.25.

Since the charge on each playe should be q = 200 µC = 200*10^-6C

Now the additional charge accumulated should be

qa = 250µC = 250*10^-6C

Now we know that

Ci = q/v

Ci = 200µC/ V

Also,

Q = q+ qa

= 9200 + 250)µC

= 450µC

Now C = 450µC /  V

So here the dielectric constant should be

= 450µC/200µC

= 2.25

Learn more about charge here; https://brainly.com/question/24140726

Answer: High voltage transmission minimizes energy losses during electricity transmission.

Explanation:

When electricity is to be transmitted over a long distance, high voltage transmission is preferred to minimize energy losses due to heat.

The higher the transmission voltage, the lower the current and the lesser the resitance in the wire and the lesser the energy lost due to heat during transmission.

High voltage transmission in kilowatts enables light weight cables to be used for long distance electricity transmission.

This explains why, even though only 110 V may be required at home by some appliances, but electricity is transmitted in kilovolts and often require a stepdown transformer

Answer:

In the statement of Newton's first law, the unbalanced force refers to that force that does not become completely balanced (or canceled) by the other individual forces. If either all the vertical forces (up and down) do not cancel each other and/or all horizontal forces do not cancel each other, then an unbalanced force exists. The existence of an unbalanced force for a given situation can be quickly realized by looking at the free-body diagram for that situation. Free-body diagrams for three situations are shown below. Note that the actual magnitudes of the individual forces are indicated on the diagram.

In each of the above situations, there is an unbalanced force. It is commonly said that in each situation there is a net force acting upon the object. The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out. At this point, the rules for summing vectors (such as force vectors) will be kept relatively simple. Observe the following examples of summing two forces:

Observe in the diagram above that a downward vector will provide a partial or full cancellation of an upward vector. And a leftward vector will provide a partial or full cancellation of a rightward vector. The addition of force vectors can be done in the same manner in order to determine the net force (i.e., the vector sum of all the individual forces). Consider the three situations below in which the net force is determined by summing the individual force vectors that are acting upon the objects.

Distance= Time×Speed

= 1800×1.5

= 2700 m

I am not sure it's right. the question itself is confusing.

Answer:

The value is [tex]t_t =41.12 \ s [/tex]

Explanation:

From the question we are told that

The maximum acceleration is [tex]a_m = 2.10 \ m/s^2[/tex]

The maximum velocity is [tex]v_m = 10.3 \ m/s[/tex]

The height of the observatory floor is [tex]H = 373 \ m[/tex]

Generally at the initial state of the motion the elevator accelerates from rest with a maximum acceleration [tex]a_m = 2.10 \ m/s^2[/tex] to attain a maximum velocity [tex]a_m = 2.10 \ m/s^2[/tex] at a time period of [tex]t_a[/tex]

Then it continues with the maximum velocity for a time period of [tex]t_b[/tex]

and finally decelerates with a maximum deceleration [tex] a_m = -2.10 \ m/s^2[/tex] to come to rest at a time period of [tex]t_c[/tex]

Generally the velocity of the elevator for the first part of the motion is

[tex]v = u + at_a[/tex]

given that it started from rest u = 0 m/s

=> [tex]10.3 = 2.10t_a[/tex]

=> [tex]t_a = 4.90 \ s [/tex]

Generally from kinematic equations the distance covered during the first part of the motion is

[tex]S_1 = ut_a + \frac{1}{2} at_a^2[/tex]

=> [tex]S_1 = 0 + \frac{1}{2} (2.10 ) (4.90)^2[/tex]

=> [tex]S_1 = 25.21 \ m [/tex]

Generally the distance covered during the second part of the elevators motion is mathematically represented as

[tex]S_2 = v_m * t_2[/tex]

[tex]S_2 = 10.3 t_b[/tex]

Generally given that the first part of the elevator motion is similar to the third part of the elevator motion it implies that the distance covered is similar

i.e [tex]S_1 = S_3 = 25.21 \ m[/tex]

The total distance covered is mathematically repented as

[tex]H = S_1 + S_2 + S_3[/tex]

=> [tex]373 = 25.21 + 10.3 t_b + 25.21 [/tex]

=> [tex]t_2 = 31.32 \ s [/tex]

Generally the minimum total time taken is

[tex]t_t = t_a + t_b + t_c[/tex]

Generally [tex]t_a = t_c = 4.90 \ s[/tex]

So

     [tex]t_t  =4.90+ 31.32 +4.90[/tex]

=> [tex]t_t =41.12 \ s [/tex]

Answer:

The minimum acceleration his partner will need in order to catch Bond before he hits the water is 25.688 meters per square second.

Explanation:

Let be the initial position of the sppedboat the reference location, James Bond jumps horizontally and experiments a parabolic motion, which consists in a horizontal motion at constant velocity and an vertical uniform accelerated motion due to gravity.

Speedboat must accelerate from rest and get Bond at right location and instant. Now we describe the appropriate equations of motion for James Bond and for the speedboat:

James Bond

[tex]x_{B} = x_{B,o} + v_{B,o,x} \cdot t[/tex]

[tex]y_{B} = y_{B,o} + v_{B, o, y}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]

Speedboat

[tex]x_{S} = x_{S,o}+v_{S,o}\cdot t + \frac{1}{2}\cdot a_{S}\cdot t^{2}[/tex]

Where:

[tex]x_{B,o}[/tex], [tex]x_{S,o}[/tex] - Initial horizontal positions for James Bond and the speedboat, measured in meters.

[tex]x_{B}[/tex], [tex]x_{S}[/tex] - Current horizontal positions for James Bond and the speedboat, measured in meters.

[tex]v_{B,o,x}[/tex], [tex]v_{S,o}[/tex] - Initial horizontal velocities of James Bond and the speedboat, measured in meters per second.

[tex]v_{B,o, y}[/tex] - Initial vertical velocity of James Bond, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]a_{S}[/tex] - Acceleration of the speedboat, measured in meters per square second.

We must observe the following conditions such as we may calculate the minimum acceleration of the speedboat:

1) [tex]x_{B} = x_{S}[/tex]

2) [tex]y_{B} = 0\,m[/tex]

Now, if we know that [tex]x_{B,o} = 60\,m[/tex], [tex]x_{S,o} = 0\,m[/tex], [tex]y_{B, o} = 18\,m[/tex], [tex]y_{B} = 0\,m[/tex], [tex]v_{B,o, x} = -6.7\,\frac{m}{s}[/tex], [tex]v_{S,o} = 0\,\frac{m}{s}[/tex], [tex]v_{B,o,y} = 0\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the resulting system of equations is:

[tex]60\,m +\left(-6.7\,\frac{m}{s} \right)\cdot t = 0\,m +\left(0\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot a_{S}\cdot t^{2}[/tex]

[tex]\frac{1}{2}\cdot a_{S}\cdot t^{2} +\left(6.7\,\frac{m}{s} \right)\cdot t -60\,m = 0[/tex] (Eq. 1)

[tex]0\,m = 18\,m +\left(0\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}+18\,m = 0[/tex] (Eq. 2)

At first we get the time from (Eq. 2) by using the Quadratic Formula:

[tex]t_{1} \approx 1.916\,s[/tex] and [tex]t_{2}\approx -1.916\,s[/tex]

Only the first root is physically reasonable. ([tex]t\approx 1.916\,s[/tex])

Then, we substitute time in (Eq. 1) and clear the acceleration of the speedboat:

[tex]\frac{1}{2}\cdot a_{S}\cdot (1.916\,s)^{2} +\left(6.7\,\frac{m}{s} \right)\cdot (1.916\,s) -60\,m = 0[/tex]

[tex]1.836\cdot a_{S}-47.163\,m = 0[/tex]

[tex]a_{S} = 25.688\,\frac{m}{s^{2}}[/tex]

The minimum acceleration his partner will need in order to catch Bond before he hits the water is 25.688 meters per square second.

Answer:

Let there be vectors A and B  where B = -A

Then A + B = 0