A monkey has a bit of a heavy for on the gas pedal. As soon as the light turns green the monkey pushes the gas pedal to the floor and accelerates from 0 mph to 60 mph (0 m/s - 28.82 m/s) in 4.2 seconds, calculate the cars acceleration.

Answer:

Explanation:

The attachments show the labeling of the network nodes. We found it convenient to do this so we could identify the specific nodes any given branch was attached to. Then the connections for each resistor were identified, and the list sorted to make it easier to see parallel connections.

Resistors 1 and 2 are in parallel

Resistors 5, 6, 8, 9 are in parallel

Resistors 4, 7, 11 are in parallel

No resistor has a parallel connection to resistor 3.

Answer:

The value of m = 3.735 and the value of n = 7.

Explanation:

The equation for the momentum of the train is,

                                          P = mv  

Here, m is the mass of the train and v is the speed of the train.

Substitute [tex]4.5 \times {10^5}{\rm{ kg}} , 8.3 \times {10^1}{\rm{ m/s}}[/tex] for m and v respectively in above equation.

[tex]\begin{array}{c}\\P = \left( {4.5 \times {{10}^5}{\rm{ kg}}} \right)\left( {8.3 \times {{10}^1}{\rm{ m/s}}} \right)\\\\ = 37.35 \times {10^6}{\rm{ kg}} \cdot {\rm{m/s}}\left( {\frac{{{{10}^1}{\rm{kg}} \cdot {\rm{m/s}}}}{{10\,{\rm{kg}} \cdot {\rm{m/s}}}}} \right)\\\\ = 3.735 \times {10^7}{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}[/tex]  

According to the scientific notation, here the value of m is 3.735 and the value of n is 7 in the final answer of the momentum.

                        The value of m = 3.735 and the value of n = 7.

Answer:

Explanation:

check the attachment for the diagram of the set up for proper clarification.

From the diagram, the man's distance from the pole is labelled AB.

To get AB =, we need to get side CB first which is equal to ED i.e ED = CB.

From ΔCDE, opposite side = 45 feet and the adjacent is ED.

USing TOA, according to trig function SOH, CAH, TOA;

tan 16° = opp/adj = BD/ED

tan 16° = 45/ED

ED = 45/tan 16°

ED = 156.93 feet

Since ED = CB, hence CB = 156.93 feet

From ΔABC, adj = CB and hypotenuse = AB. According to CAH;

cos 12° = adj/hyp = CB/AB

cos 12° =  156.93/AB

AB =  156.93/cos 12°

AB = 160.44 feet

Hence the man's distance from the pole is 160.44 feet

Answer:

d) electrical repulsion

Explanation:

When you stand on a floor, the two different bodies i.e your shoes and the floor on which you stand, are in a very close contact, the electrons that surround your shoes and that of the floor exert a repulsive force on each other, also known as Coulomb's force of repulsion or  electrical repulsive force.

This electrical repulsive force prevents you from falling through the floor.

Answer:

Distance is 13m and Displacement is 9m

Explanation:

the endocrine system releases hormones, and hormones impact the systems that cause changes in our behavior, including changes in what we might call biological motivations.

Answer:

The car's average acceleration points due west.

Explanation:

Resolving the acceleration will give a resultant due west

Explanation:

(m2-m1)/t

25-0/5

25/5

5m/s

Answer: The absolute value of a number is symbolized by two vertical lines as 3 or -3

Please pick me as brainliest please

Answer: the magnetic of a real number without regard to its sign

Explanation: In mathematics, the absolute value or modulus of a real number x, denoted |x| or, is the non-negative value of x without regard to its sign. Namely, |x| = x if x is positive, and |x| = −x if x is negative, and |0| = 0. For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3

Use the formula,

[tex]\Delta x=v_it+\dfrac12at^2[/tex]

where [tex]\Delta x[/tex] is the cart's displacement (from the origin), [tex]v_i[/tex] is its initial speed, [tex]a[/tex] is its acceleration, and [tex]t[/tex] is time.

[tex]\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2[/tex]

[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]

Alternatively, since acceleration is constant, we have

[tex]\dfrac{v_f+v_i}2=\dfrac{\Delta x}t[/tex]

That is, we have these two equivalent expressions for average velocity, where [tex]v_f[/tex] is the cart's final velocity. Solve for [tex]\Delta x[/tex]:

[tex]\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}[/tex]

[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]

The required value is required in SI units.

The required answer is [tex]1.94\times10^{-10}\ \text{kg m}^2/\text{s}^2[/tex]

The SI unit of mass, length and time is kg, m and s respectively.

In order to convert one unit into another it has to be multiplied or divided by the conversion factors.

A definite magnitude which has some quantity which is defined by convention or law is called a unit.

The conversion factors are

[tex]1\ \text{g}=10^{-3}\ \text{kg}[/tex]

[tex]1\ \text{cm}=10^{-2}\ \text{m}[/tex]

[tex]1\ \text{cm}^2=10^{-4}\ \text{m}^2[/tex]

1 min = 60 s

[tex]1\ \text{min}^2=60\times60\ \text{s}^2[/tex]

So,

[tex]7\ \text{g cm}^2/\text{min}^2=7\times \dfrac{10^{-3}\times 10^{-4}}{60\times 60}\\ =1.94\times10^{-10}\ \text{kg m}^2/\text{s}^2[/tex]

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Answer: A velocity-time graph shows how velocity changes over time

Explanation:

Answer:velocity

Explanation:

The car will move in a speed of 45 meter per second

Answer: The car will move in a speed of 45 meter per second

Explanation:

Answer:

Answer: potassium (K) germanium (Ge) selenium (Se)

Explanation:

I just took the test, the farther to the right on the periodic table you go the less conductive it gets.

Answer:

Answer: potassium (K) germanium (Ge) selenium (Se)

Explanation:

The most conductive to least conductive is from left to right

Answer:

When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.

Answer:

The acceleration of the sled is [tex]4.6\ m/s^2[/tex].

Explanation:

It is given that,

Initial speed of sled is 0 because it was at rest.

It is placed at an angle of 28° on a frictionless hill.

We need to find the acceleration of the sled. It is placed at an incline. It means that the acceleration of the sled is given by :

[tex]a=g\sin\theta\\\\a=9.8\times \sin(28)\\\\a=4.6\ m/s^2[/tex]

So, the acceleration of the sled is [tex]4.6\ m/s^2[/tex].

Answer: 13.2678 kg

Explanation: Well, 13.2678 kg is 29.25 in mass and kg.

Answer:

1)   I_ pendulum = 2.3159 kg m² , 2)  I_pendulum = 24.683 kg m²

Explanation:

In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum

Sphere

They indicate the density of the sphere roh = 37800 kg / m³ and its radius

r = 5 cm = 0.05 m

we use the definition of density

               ρ = M / V

               M = ρ V

the volume of a sphere is

                V = 4/3 π r³

we substitute

              M = ρ 4/3 π r³

we calculate

              M = 37800  4/3 π 0.05³

              M = 19,792 kg

Bar

the density is ρ = 32800 kg / m³ and its dimensions are 1 m,

0.8 cm = 0.0008 m and 4cm = 0.04 m

The volume of the bar is

               V = l w h

              m = ρ l w h

we calculate

              m = 32800 (1   0.008   0.04)

              m = 10.496 kg

Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is

               r_cm = 1 / M (M r_sphere + m r_bar)

               M = 19,792 + 10,496 = 30,288 kg

               r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))

               r_cm = 1 / 30,288 (5,248 + 20,7816)

               r_cm = 0.859 m

This is the center of mass of the pendulum.

1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:

Sphere I = 2/5 M r2

Bar I = 1/12 m L2

parallel axes theorem

                  I = I_cm + m D²

where m is the mass of the body and D is the distance from the body to the axis of rotation

Sphere

      m = 19,792 ka

the distance D is

                D = 1.05 -0.85

                D = 0.2 m

we calculate

               I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168

               I_sphere = 0.811472 kg m²

Bar

m = 10.496 kg

distance D

             D = 0.85 - 0.5

             D = 0.35 m

              I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576

              I_bar = 1.5044 kg m²

The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts

              I_pendulum = I_sphere + I_bar

              I_pendulum = 0.811472 +1.5044

              I_ pendulum = 2.3159 kg m²

this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m

2) The moment is requested with respect to the pivot point at r = 0 m

Sphere

        D = 1.05 m

         I_sphere = 2/5 M r2 + M D2

        I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82

        I_sphere = 21.84 kg m²

Bar

D = 0.5 m

      I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624

      I_bar = 2,84266 kg m 2

The pendulum moment of inertia is

       I_pendulum = 21.84 +2.843

       I_pendulum = 24.683 kg m²

This moment of inertia is about the turning point at r = 0 m

Answer:

ANSWER

A = 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

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Answer:

[tex]here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \\ t = .2sec \\ omega = \frac{2\pi}{t} \\ = \frac{2\pi}{.2 } = 10\pi \frac{rad}{s} \\ we \: want \: find \: a \: and \: v \\ we \: know \: that \: a = - {omega}^{2} x \\ v = omega \sqrt{ {r}^{2} - {x}^{2} } \\(1)x = .05m \\ a = - {10\pi}^{2} \times .05 = - 5 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.05}^{2} } = 0 \\ 2)x = 3cm = .03m \\ a = {(10\pi)}^{2} \times .03 = - 3 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.03}^{2} } = 10\pi \times .04 = .4\pi \frac{m}{s} \\ 3)x = 0 \\ a = - {(10\pi)}^{2} \times 0 = 0 \\ v = 10\pi \times \sqrt{ {.05}^{2} - {0}^{2} } = - 10\pi \times .05 = .5\pi \frac{m}{s} \\ thank \: you[/tex]

Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

[tex]y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}[/tex]

(1)

[tex]y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s[/tex]

(2)

[tex]y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s[/tex]

(3)

[tex]y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s[/tex]

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

(a)The average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

The total displacement traveled by an object divided by the total time taken is the average velocity.

Average velocity is given by;

[tex]\rm y(t_2,t_1)=\frac{y(t_2)-y(t_1)}{t_2-t_1} \\\\[/tex]

The average velocity for case 1;

[tex]\rm y(2.1,2)=\frac{(235-16\times 2\times 1^2)-(235-16 \times 2^2)}{2.1-2} \\\\ \rm y(2.1,2)=\-65.6 \ m/sec[/tex]

The average velocity for case 2;

[tex]\rm y(2.015,2)=\frac{(235-16\times 2.05^2)-(235-16 \times 2^2)}{2.05-2} \\\\ \rm y(2.1,2)=\-64.8 \ m/sec[/tex]

The average velocity for case 3;

[tex]\rm y(2.01,2)=\frac{(235-16\times 2.01^2)-(235-16 \times 2^2)}{2.01-2} \\\\ \rm y(2.1,2)=\-64.16 \ m/sec[/tex]

Hence the average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

(b) The instantaneous velocity will be -96 m/s.

The given equation in the problem is;

[tex]\rm y = 235 - 16t^2[/tex]

The instantaneous velocity is given as;

[tex]v = \frac{dy}{dt} \\\\ \frac{dy}{dt} = -32t\\\\ t = 3 s\\\\ v = -32\times 3\\\\ v = -96 m/s[/tex]

Hence the instantaneous velocity will be -96 m/s.

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Given :

The acceleration of a piston is given by :

[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]

To Find :

The amplitude of the acceleration of piston .

Solution :

Now , acceleration is :

[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]

Now , amplitude is maximum of any function .

So , maximum value of [tex]cos\ \theta[/tex] is equal to 1 .

Therefore , amplitude of the acceleration of piston is :

[tex]A=r\omega^2[/tex]

Hence , this is the required solution .

Answer:

The darts hit the same general area of the board.

Answer: C. The darts hit the same general area of the board.

Explanation: The results of a dart game were precise but not accurate. The accepted value of the game was the center of the dartboard.

Answer: The amount of matter is higher in a bus than a ball.

Explanation:

Answer:C: The gravity on the moon is less than the gravity on Earth.

Explanation:

Answer:

B, and D

Explanation:

The two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.

The water cycle is defined as a cycle of events that involves precipitation as rain and snow, drainage in streams and rivers, and return to the atmosphere by evaporation and transpiration. Water moves through this cycle between the earth's oceans, atmosphere, and land.

It can also be defined as the route all water takes as it travels around Earth in various conditions.

There are basically six stages of water cycle.

Thus, the two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.

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Answer:

The new water temperature is 26.4 °C

Explanation:

Given;

mass of copper, [tex]M_{cu}[/tex] = 26 g = 0.026 kg

temperature of copper, t = 300 °C

volume of water, V = 120 mL = 0.12 L

temperature of water, t = 21 °C

density of water, ρ = 1 kg/L

mass of water = density x volume

mass of water = (1 kg/L) x 0.12 L = 0.12 kg

heat lost by copper = heat gained by water

Both copper and water reach final temperature, T

Heat gained by water, [tex]Q_w[/tex] = [tex]m_w[/tex]cΔθ = [tex]m_w C(T - t)[/tex]

[tex]Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)[/tex]

Heat lost by copper is given by;

[tex]Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)[/tex]

[tex]Q_{cu} = Q_w[/tex]

504(T- 21) = 10.01(300 - T)

504 T - 10584 = 3003 - 10.01 T

504 T + 10.01 T= 3003 + 10584

514.01 T = 13587

T = (13587) / 514.01

T = 26.4 °C

Therefore, the new water temperature is 26.4 °C

The new water temperature using the given parameters is;

26.49°C

We are given;

Mass of copper; m_cu = 26 g = 0.026 kg

Temperature; T_cu = 300 °C

Volume of water; V = 120 mL = 0.12 L

Temperature of water, T_w = 21 °C

Density of water; ρ = 1 kg/L

Let us find the mass of water from the formula;

m_w = ρ × V

m_w = 1 × 0.12

m_w = 0.12 kg

Now, from the principle of conservation of energy, we can say that the total heat lost by a hot body is equal to the total heat gained by a cold body.

The hot body here is copper while the cold body is water. Thus;

Heat lost by copper = Heat gained by water

Formula for heat lost by copper is;

Q_cu = m_cu * c_cu * (T_f - T_cu)

Formula for heat gained by water is;

Q_w = m_w * c_w * (T_f - T_w)

Where;

T_f is final temperature reached by copper and water

c_cu is specific heat capacity of copper = 385 J/kg.°C

c_w is specific heat capacity of water = 4200 J/kg.°C

Thus;

Q_cu = 0.026 × 385 × (300 - T_f)

Q_cu = 3030 - 10.01T_f

similarly;

Q_w = 0.12 × 4200 × (T_f - 21)

Q_w = 504T_f - 10584

Thus;

3030 - 10.01T_f = 504T_f - 10584

3030 + 10584 = 504T_f + 10.01T_f

13614 = 514.01T_f

T_f = 13614/514.01

T_f = 26.49°C

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Answer:

C An ideal isotropic antenna is a theoretical antenna that does not exist in practice, but is useful in explaining power density and unguided EM signal attenuation

Explanation:

Because practically speaking there is no ideal isotropic antenna it is only imaginary radiating in all directions and use as an arbitrary point for antenna gain