A wave can be best defined as

P.S pls help

Answer:

A. [tex]2\cdot t[/tex]

Explanation:

The SUV accelerates uniformly. In this case, speed ([tex]v[/tex]) is directly proportional to time ([tex]t[/tex]). That is:

[tex]v \propto t[/tex] (1)

[tex]v = k\cdot t[/tex]

Where [tex]k[/tex] is the proportionality constant.

Then, we eliminate this constant by creating the following relationship:

[tex]\frac{v_{1}}{t_{1}} = \frac{v_{2}}{t_{2}}[/tex] (2)

If we know that [tex]v_{1} = v[/tex], [tex]t_{1} = t[/tex] and [tex]v_{2} = 2\cdot v[/tex], then the time interval for the SUV to accelerate from rest to a speed [tex]2\cdot v[/tex] is:

[tex]\frac{v}{t} = \frac{2\cdot v}{t_{2} }[/tex]

[tex]t_{2} = 2\cdot t[/tex]

Hence, correct answer is A.

Answer:

a= 0

Explanation:

       [tex]F_{net} = m*a = 0 (1)[/tex]

      ⇒ a = 0

Answer:

63°

Explanation:

90-27 =63

I am not completely sure

Answer:

3.6 arcsec

Explanation:

 angular diameter = diameter / distance

diameter is constant

so angular diameter ∝ 1 / distance

angular diameter = k / distance

For first case ,

18 = k / .5

for second case let angular diameter be D .

D = k / 2.5

dividing ,

D / 18 = .5 / 2.5 = 1 / 5

D = 18 / 5 = 3.6 arcsec

3.6 arcsec is the answer .

Answer:

c

Explanation:

Answer: C

Explanation:

Answer:

The net acceleration of the SUV is 0.429 meters per square second due west.

Explanation:

Statement is incomplete. Description is presented below:

A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?

According to Newton's Laws of Motion, the SUV will accelerate if and only if net acceleration is different of zero. Let suppose as positive the direction of driving force ([tex]F[/tex]), measured in newtons:

[tex]\Sigma F = F - R -f = F_{net}[/tex] (1)

Where:

[tex]R[/tex] - Resistance force, measured in newtons.

[tex]f[/tex] - Wind force, measured in newtons.

[tex]F_{net}[/tex] - SUV net force, measured in newtons.

If we know that [tex]F = 2500\,N[/tex], [tex]R = 500\,N[/tex] and [tex]f = 500\,N[/tex], then net force experimented by the SUV is:

[tex]F_{net} = 2500\,N-500\,N-500\,N[/tex]

[tex]F_{net} = 1500\,N[/tex]

The car has acceleration.

By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:

[tex]a_{net} = \frac{F_{net}}{m}[/tex] (2)

Where [tex]m[/tex] is the mass of the SUV, measured in kilograms.

If we know that [tex]F_{net} = 1500\,N[/tex] and [tex]m = 3500\,kg[/tex], then the net acceleration of the car is:

[tex]a_{net} = \frac{1500\,N}{3500\,kg}[/tex]

[tex]a_{net} = 0.429\,\frac{m}{s^{2}}[/tex]

The net acceleration of the SUV is 0.429 meters per square second due west.

Answer:

If it accelerates at 20 m/s2 for a period of 22 seconds, what is its final velocity? ... How fast is the ball falling after 5 seconds? v = v0 + gt v = 0 + 10(5) v = 50 m/s. 4. ... + ½ 2.5(15)2 x = 281 m. 5. What is the total displacement of the car in question 2? ... 8. A base jumper falls until he reaches a speed of 200 m/s

Explanation:

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Answer:

[tex]c=0.45\ J/g^{\circ} C[/tex]

Explanation:

Given that,

A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at  30.00°C). The final temperature of the system is 40.22°C.

We need to find the specific heat of iron.

It can be calculated as:

Cooler water gains = hot metal loses

mc∆T = - mc∆T

Put all the values,

[tex]200g(4.184\ J/g^{\circ} C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C[/tex]

So, the specific heat of iron is [tex]0.45\ J/g^{\circ} C[/tex]

Answer:

A: 1,212 kg

Explanation:

Answer:

At zero kelvin (minus 273 degrees Celsius) the particles stop moving and all disorder disappears. Thus, nothing can be colder than absolute zero on the Kelvin scale. Physicists have now created an atomic gas in the laboratory that nonetheless has negative Kelvin values.

Explanation:

Answer:

a) the rate at which 235U is fissioned is 2895.9 grams per day

b) the rate at which 235U is consumed is 3385.3071 gram/day

Explanation:

Given the data in the question;

a)

designed operation thermal power = 2758 MW.

we know that, the burn up rate fission rate of 235U is 1.05 grams per MW-day.

so, the rate at which 235U is fissioned will be;

⇒ 2758 × 1.05 = 2895.9 grams per day

Therefore, the rate at which 235U is fissioned is 2895.9 grams per day

b)

Consumption rate is given as;

Cr = 1.05 × ( 1 + ∝ )P gram/day

where ∝ is 0.169 for U-235

so,

Cr = 1.05 × ( 1 + 0.169 )2758 gram/day

Cr = 1.05 × 1.169 × 2758 gram/day

Cr = 3385.3071 gram/day

Therefore, the rate at which 235U is consumed is 3385.3071 gram/day

Answer:

d=20m/sx60s=1200m=1200/1000Km=1.2km

Explanation:

Answer:

A

Explanation:

In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.

Hope this helps!

Answer:

The answer is below

Explanation:

The resistance of a wire is directly proportional to the length of the wire and inversely proportional to its area. The resistance (R) is given by:

[tex]R=\frac{\rho L}{A}\\\\where\ L=length \ of\ wire,A=cross\ sectional\ area, \rho=resistivity\ of\ wire.[/tex]

Let us assume that all the wires have the same resistivity.

a) Wire of Length L and area A

[tex]R_1=\frac{\rho L}{A}[/tex]

b) Wire of Length 2L and area A

[tex]R_2=\frac{\rho *2L}{A}=2R_1[/tex]

C) Wire of Length L and area 2A

[tex]R_3=\frac{\rho L}{2A}=\frac{1}{2}R_1[/tex]

Therefore the wire of least resistance is R3 and R2 has the highest resistivity.

R₃ < R₁ < R₂

Therefore, the ranking of the wires from most current (least resistance) to least current (most resistance) is:

R₃ < R₁ < R₂

Answer:

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Complete Question:

Gauss's law:

Group of answer choices

A. can always be used to calculate the electric field.

B. relates the electric field throughout space to the charges distributed through that space.

C. only applies to point charges.

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

E. relates the surface charge density to the electric field.

Answer:

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

Where;

ϕ is the electric flux.

Q represents the total charge in an enclosed surface.

ε0 is the electric constant.

are you a dmbss? the bulb and wire must be connected to both end

Answer:

a.432

B. 562

C.402

D. 316

A

Explanation:

Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied. Work was done by the workman will be 432 Nm.

What is work?

Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative.it depend on the diection of body displaced . if the body is displaced in thw same direction of force it will be positive .

while if the displacement is in the opposite direction of force applied the work will be negative work . if their is no displacement of the body the work done will b e zero.

As given in question wall is stationary diplacement will be zero so in that work will also z

The given data in the problem is;

x is the displacement in the horizontal direection=22m

y is the displacement in the verical direection=9.6m

W is the weight of lumber = 45 N

[tex]\rm W=F\times y[/tex]

[tex]\rm W=F\times y \\\\\rm W=45\times 9.6\\\\ \rm W=432\;Nm[/tex]

Hence Work was done by the workman will be 432 Nm.

To learn more about the work refer to the link ;

https://brainly.com/question/3902440

Answer:

a

Explanation:

Answer:

With or without.

Explanation:

According to the National Electrical Code (NEC), here the air conditioner disconnecting means is not within sight from the equipment, the provision for locking or adding a lock to the disconnecting means shall be on the switch or circuit breaker and remain in place with or without the lock installed. Thus, this is in accordance with section 110.25 of the National Electrical Code (NEC).

Answer:

$84

Explanation:

The coefficient of performance (COP) show the relationship between the power (kW) output of the heat pump and the power (kW) input to the compressor.

The heater consumed by the heater is 1200 kWh.

For a heat pump with a COP of 2.4, the electric input needed to produce an output of 1200 kWh is:

Electric input to heat pump = 1200 / 2.4 = 500 kWh

That means that supplying a heat pump with 500 kWh produces an output of 1200 kWh

The amount of power saved = power consumed by heater - power consumed by heat pump = 1200 - 500 = 700 kWh

Money saved = $0.12/kWh * 700 kWh = $84

The amount of money the homeowner would have saved during the month is $84.

Given the following data:

To determine the amount of money the homeowner would have saved during the month:

In Science, the coefficient of performance (COP) is a mathematical expression that is typically used to show the relationship between the power output (kilowatts) of a heater and the power input (kilowatts) to the compressor.

Mathematically, the coefficient of performance (COP) is given by the formula:

[tex]COP = \frac{E_o}{E_i}[/tex]

Where:

First of all, we would determine the energy input of the heaters.

[tex]E_i = \frac{E_o}{COP} \\\\E_i=\frac{1200}{2.4}[/tex]

Energy input = 500 kWh

For power saved:

[tex]Power = 1200 -500[/tex]

Power = 700 kWh

For the amount of money saved while using these heaters:

[tex]Money = 0.12 \times 700[/tex]

Amount of money = $84

Read more: https://brainly.com/question/22599382

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          [tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_i m_i[/tex]

in this case we have two bodies

          x_{cm} = [tex]\frac{1}{M}[/tex] (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = [tex]\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )[/tex]

          x_{cm} = [tex]\frac{1}{M} ( m_1 v_1 + m_2 v_2 )[/tex]

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = [tex]\frac{m_1}{m_1+m_2}[/tex]  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         [tex]v_{cm o}[/tex] = [tex]\frac{1}{m_1 +m_2}[/tex] (m₁ vo)

final

         [tex]v_{cm f}[/tex] = [tex]\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )[/tex] ( v) = v

         v_{cm f} =  [tex]\frac{m_1}{M^2} v_o[/tex]

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = [tex]\frac{1}{M} = \frac{1}{m_1 +m_2}[/tex]

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = [tex]\frac{m_1 -m_2}{m1 +m2 } v_o[/tex]

             v_{2f} = [tex]\frac{2 m_1}{m-1+m_2} vo[/tex]

now let's find the velocity of the center of mass

initial

          [tex]v_{cm o}[/tex] = [tex]\frac{1}{M}[/tex] m₁ v₀

final

          [tex]v_{cm f}[/tex] = [tex]\frac{1}{M}[/tex]  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = [tex]\frac{1}{M}[/tex] [  [tex]m_1 \frac{m_2}{M}[/tex] + [tex]m_2 \frac{2 m_1}{M}[/tex] ] v₀

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = [tex]\frac{1}{M}[/tex] M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

Answer:

el conductor

Explanation:

gracias por los puntitoss

Answer:

conductor

Explanation:

Answer:−4.05

Explanation:

Answer:

she should write about how big is it and what the sun looks and how far away is it from earth.

Answer:

beacuse it isnt on

Explanation:

Explanation:

The wires isn't in a circulation, the negative side must be connected to the wire