A man slides on snow without friction starting at 8.96m/s at the top of an inclined plane with height 8.21m. What is his speed at the bottom of a plane?​

Answer:

yes

Explanation:

becuse it is an even  that is 2 number then 20

Answer:

Yes

Explanation:

It is because 22 is greater than 20

Answer:

The description is outlined throughout the clarification section following, and according to the given word.

Explanation:

Answer:

Explanation:

BAHAHAHAH karely omg HAHAHA un idke

Answer:

P=2940 Pa

Explanation:

Given that,

Side of a cube, a = 12 cm

It is submerged to  depth of 30 cm

The density of water, d = 1000 kg/m^3

We need to find the pressure at the bottom surface of the cube. We know that the pressure exerted is given by :

[tex]P=d gh\\\\P=1000\ kg/m^3\times 9.8\ m/s^2\times 0.3\ m\\\\P=2940\ Pa[/tex]

So, 2940 Pa of pressure is exerted at the bottom surface of the cube due to water.​

Answer:

Diagram B.

Explanation:

We'll begin by defining potential energy.

This is given below:

Potential energy is the energy stored in an object due to its position. Mathematically, it is expressed as:

P.E = m × g × h

Where:

P.E is the potential energy.

m is the mass of the object.

g is the acceleration due to gravity.

h is the height to which the object is located.

From:

P.E = m × g × h

We can say that the potential energy (P.E) is directly proportional to the height (h) to which the object is located. This implies that as the height increase, the potential energy of the object will also increase and as the height decrease, the potential energy of the object will also decrease.

Now, considering the question given above, we can see that the height of the object is greater in diagram B. Therefore, The potential energy of the object is greater in diagram B.

Answer:

The answer s B

Explanation:

Answer:

It can be either, depending on how much and how you add the electrons

Answer:

it's a Bugatti

Explanation:

I seen pictures and videos of one for what they look like from the inside or the outside

Answer:

the mug will moveforwards

Answer:

λ₀= 495.88 nm

Explanation:

To analyze this constructive interference interference experiment by reflection, let's look at two important aspects:

* when a ray of light passes from a medium with a lower index, they refact to another medium with a higher index, the reflected ray has a phase difference of pyres

* When a beam penetrates a material medium, the wavelength of the radiation changes according to the refractive index of the material.

       λₙ = λ₀ / n

when we introduce these aspects in the expression of contributory interference, it remains

        2 d sin θ = (m + ½) λ₀ / n

In general, reflection phenomena are measured at an almost normal angle, whereby θ = π/2  and sin θ = 1

        2 d = (m +1/2) λ₀/ n

         2n d = (m + ½) λ₀

Let's apply this expression to our case

         d = (m + ½) λ₀ / 2n

Suppose we measure on the first interference, this is m = 0

         d = ½ λ₀ / 2n

let's calculate

         d = ½ 496 10⁻⁹ / (2 2.30)

         d = 53.9 10-9 m  

This is the thickness of the glass, the next wavelength that gives constructive interference is

          λ₀ = 2 n d / (m + ½)

let's calculate

          λ₀ = 2 2.3 5.39 10-8 / (1 + ½)

          λ₀= 4.9588 10-7 m

          λ₀= 495.88 nm

Answer:

a

[tex]T_2 =  276.1 \ K[/tex]

b

  [tex]V_2  = 2.13 *10^{3} \ m^3 [/tex]

c

   [tex]\Delta U =  -1.25 *10^{7} \  J  [/tex]

Explanation:

From the question we are told that

  The  volume of the balloon is  [tex]V  =  2.00 * 10^3 \ m^3[/tex]

   The  pressure of  helium is  [tex]P_1 =   1.00 \ atm=  1.0 *10^{5} \  Pa [/tex]

   The  initial  temperature is  [tex]T_1   =  15.0^oC =  288 \  K [/tex]

   The  pressure of  atmosphere is [tex]P_a  =  0.900 \ atm[/tex]

 Generally the equation representing the  adiabatic process is mathematically represented as

      [tex]P_1 V_1 ^{\gamma }=  P_2 V_2 ^{\gamma }[/tex]

=>    [tex]V_2 ^ {\gamma } =  \frac{ V_1 ^{\gamma } *  P_1 }{P_2}[/tex]

Generally [tex]\gamma[/tex] is a constant with value [tex]\gamma  =\frac{5}{3}[/tex] for an ideal  gas

So  

     [tex]V_2 ^ {\frac{5}{3} } =  \frac{ ( 2.0 *10^{3}) ^{ \frac{5}{3}  } *   1.00 }{0.900}[/tex]

        [tex]V_2  =  (\sqrt[5]{103.14641852} )^3[/tex]

=>      [tex]V_2  = 2.13 *10^{3} \ m^3 [/tex]

Generally the adiabatic process can also be mathematically represented as

     [tex]T_1 V_1 ^{\gamma -1 } = T_2 V_2^{\gamma -1 }[/tex]

=>  [tex]T_2  =  288 *  [\frac{2 *  10^{3}}{ 2.13 *10^{3}} ]^{ \frac{5}{3} -1 }[/tex]

=>    [tex]T_2 =  276.1 \ K[/tex]

Generally the ideal gas equation is mathematically represented as

        [tex]P_1 V_1 = nRT_1[/tex]

Here  R is the  gas constant with value  [tex]R  =  8.314\  J  /mol \cdot  K[/tex]

     [tex]n  =  \frac{P_1 V_1 }{RT _1}[/tex]

=>  [tex]n  =  \frac{1.0 *10^{5} *  2.0 *10^{3}}{8.314 * 288[/tex]

=>  [tex]n  =  84362 \  mol[/tex]

Generally change in internal energy i mathematically represented  

        [tex]\Delta U =  n C_v  \Delta  T[/tex]

Here  [tex]C_v[/tex] is  the  specific heats of gas at constant volume and the value is  [tex]C_v  =  12.47 J/mol \cdot  K[/tex]

         [tex]\Delta U =  84362 *   12.47 * [T_2 - T_1 ]  [/tex]

         [tex]\Delta U =  84362 *   12.47 * [276.1 - 288 ]  [/tex]

          [tex]\Delta U =  -1.25 *10^{7} \  J  [/tex]

That depends on the car's speed at the moment of impact, and how long it takes to stop after the impact.

It is NOT 8,000N. That answer would come from a totally incorrect application of " F = m a " . 8,000 Newtons is what it takes to accelerate that car in the first place, long before it hits the tree.

Force can be described as either a pull or a push

Now when a body with mass and acceleration hits an object, we say a collision has taken place, hence there will be transfer of energy within the system

Give data

mass = 4000kg

acceleration = 2m/s^2

We know that force is given as

Force = mass x acceleration

Substitute our given data into the expression for force

Force = 4000 x 2Force = 8000N

For more information on force kindly visit

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Answer:

In physics, mass–energy equivalence is the principle that anything having mass has an equivalent amount of energy and vice ve

Answer:

c,B,a,d i guess i m not sure it is the guessing

Answer:

1. (b) 2.5 eV

2. (c) 430000 m/s

3. (a) The maximum velocity of any ejected electrons will decrease.  

Explanation:

1)

From Einstein's Photoelectric Equation:

hc/λ = K.E + ∅

∅ = hc/λ - K.E

∅ = hc/λ - eV

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

V = Stopping Potential = 0.52 volts

h = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 405 nm = 4.05 x 10⁻⁷ m

∅ = Work Function = ?

Therefore,

∅ = (6.625 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4.05 x 10⁻⁷ m) - (1.6 x 10⁻¹⁹ C)(0.52 volts)

∅ = 4.9 x 10⁻¹⁹ J - 0.832 x 10⁻¹⁹ J

∅ = (4.075 x 10⁻¹⁹ J)(1 eV)/(1.6 x 10⁻¹⁹ J)

∅ = 2.5 eV

therefore, correct answer is"

(b) 2.5 eV

2)

K.E = (1/2)mv² = eV

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed = ?

therefore,

(1/2)(9.1 x 10⁻³¹ kg)v² = (1.6 x 10⁻¹⁹ C)(0.52 volts)

v = √(0.18 x 10¹² m²/s²)

v = 0.43 x 10⁶ m/s = 430000 m/s

Correct option is:

(c) 430000 m/s

3.

The decrease in power at constant wavelength means decrease in voltage, that results in the decrease of kinetic energy of electrons. So, correct option is:

(a) The maximum velocity of any ejected electrons will decrease.

Answer:

90

Explanation:

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Answer:

What’s the question??

Explanation:

Answer:

11,000 kg

(a) 11.2 m/s

(b) 1.6 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(2200 kg) (60.0 km/h) + m (0 km/h) = (2200 kg) (10 km/h) + m (10 km/h)

132,000 kg km/h = 22,000 kg km/h + m (10 km/h)

110,000 kg km/h = m (10 km/h)

m = 11,000 kg

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(m) (-v) + (2m) (5v) = (m) (v₁) + (2m) (v₂)

-mv + 10mv = m v₁ + 2m v₂

9mv = m (v₁ + 2 v₂)

9v = v₁ + 2 v₂

Since the collision is elastic, kinetic energy is also conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(m) (-v)² + (2m) (5v)² = m v₁² + (2m) v₂²

mv² + 50mv² = m v₁² + 2m v₂²

51mv² = m (v₁² + 2 v₂²)

51v² = v₁² + 2 v₂²

We know v = 1.60 m/s.  So the two equations are:

14.4 = v₁ + 2 v₂

130.56 = v₁² + 2 v₂²

Solve the system of equations using substitution.

130.56 = (14.4 − 2 v₂)² + 2 v₂²

130.56 = 207.36 − 57.6 v₂ + 4 v₂² + 2 v₂²

0 = 6 v₂² − 57.6 v₂ + 76.8

0 = v₂² − 9.6 v₂ + 12.8

v₂ = [ 9.6 ± √(9.6² − 4(1)(12.8)) ] / 2(1)

v₂ = 1.6 or 8

If v₂ = 1.6 m/s, then v₁ = 14.4 − 2 v₂ = 11.2 m/s.

If v₂ = 8 m/s, then v₁ = 14.4 − 2 v₂ = -1.6 m/s.

We know v₁ can't be -1.6 m/s, since that would mean puck A didn't change speeds after the collision.  Therefore, v₁ = 11.2 m/s and v₂ = 1.6 m/s.

Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the change in position.

We can use this formula to determine the launch speed of the ball. In the vertical direction, the ball has acceleration g in the downward direction, where g = 32.2 ft/s². At its maximum height, the ball has 0 vertical velocity. So we have

0² - u² = 2 (-32.2 ft/s²) (40 ft)

==>  u ≈ 50.75 ft/s

Now, the ball's height y at time t is given by

y = u t - 1/2 g t²

Find at which time t the ball covers the first 20 ft of its trajectory:

20 ft = (50.75 ft/s) t - 1/2 g t²

==>  t ≈ 0.462 s

(there is a second solution, t ≈ 2.69 s, which corresponds to the time it takes for the ball to return to this height as it's falling back down)

During this first interval, the ball's horizontal position x is

x = (20 mph) t ≈ (29.33 ft/s) t

so that at the moment the ball reaches a height of 20 ft, it will have moved

x = (29.33 ft/s) (0.462 s) ≈ 13.5 ft

to the right. (So we've take the right to be positive and the left to be negative.)

In the upper half of its trajectory, when the wind changes direction, the ball's horizontal position is given by

x = 13.5 ft - (29.33 ft/s) t

Note that t = 0 here means we now take the ball's current position to be the "initial" one. So the time we found earlier for when the ball has a height of 20 ft as it's falling back down is actually t = 2.69 s - 0.462 s ≈ 2.23 s. At this point, the ball's horizontal position is

x = 13.5 ft - (29.33 ft/s) (2.69 s - 0.462 s) ≈ -51.8 ft

or about 51.8 ft to the left of where it started.

The wind changes direction again, so that the ball's position at time t is now

x = -51.8 ft + (29.33 ft/s) t

Solving y = 0 for t gives the time when the ball reaches the ground:

0 = (50.75 ft/s) t - 1/2 g t²

==>  t ≈ 3.15 s

Again, this is the time it takes for the ball to reach the ground since it was launched, while the position function takes t = 0 to refer to the moment the ball is 20 ft above the ground. So once it hits the ground, it will be

x = -51.8 ft + (29.33 ft/s) (3.15 s - 2.69 s) ≈ -38.3 ft

which means the ball lands on the ground about 38.3 ft to the left of where it started.

Answer:

Mass

axis

distance

intertia

Answer:

a

 Solid Wire     [tex]I  =   0.01237 \  A [/tex]      

  Stranded  Wire  [tex]I_2  =   0.00978 \  A [/tex]

b

  Solid Wire   [tex]R  = 0.0149 \ \Omega [/tex]

   Stranded  Wire  [tex]R_1  = 0.0189 \ \Omega [/tex]

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  [tex]r_1  = 1.53 mm = 0.0015 \  m[/tex]

  The radius of  each strand is  [tex]r_0 =  0.306 \ mm =  0.000306 \ m[/tex]

  The current density in both wires is  [tex]J  =  1750 \  A/m^2[/tex]

Considering the first wire

     The  cross-sectional area of the first wire is

      [tex]A   = \pi  r^2[/tex]

= >  [tex]A   = 3.142 *  (0.0015)^2[/tex]

= >  [tex]A   = 7.0695 *10^{-6} \  m^2 [/tex]

Generally the current in the first wire is    

     [tex]I  =  J*A[/tex]

=>  [tex]I  =  1750*7.0695 *10^{-6}[/tex]

=>  [tex]I  =   0.01237 \  A [/tex]

Considering the second wire  wire

The  cross-sectional area of the second wire is

     [tex]A_1  =  19 *  \pi r^2[/tex]

=>     [tex]A_1  =  19 *3.142 *  (0.000306)^2[/tex]

=>  [tex]A_1  =  5.5899 *10^{-6} \  m^2[/tex]

Generally the current is  

      [tex]I_2  =  J  *  A_1[/tex]

=>    [tex]I_2  =   1750  *  5.5899 *10^{-6} [/tex]

=>    [tex]I_2  =   0.00978 \  A [/tex]

Considering question two  

 From the question we are told that

     Resistivity is  [tex]\rho  =  1.69* 10^{-8} \Omega \cdot m[/tex]

     The  length of each wire  is  [tex]l =  6.25 \  m[/tex]

Generally the resistance of the first wire is mathematically represented as

    [tex]R  =  \frac{\rho *  l  }{A}[/tex]

=> [tex]R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }[/tex]

=> [tex]R  = 0.0149 \ \Omega [/tex]

Generally the resistance of the first wire is mathematically represented as

    [tex]R_1  =  \frac{\rho *  l  }{A_1}[/tex]

=> [tex]R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }[/tex]

=> [tex]R_1  = 0.0189 \ \Omega [/tex]

Answer:

B . a heat transfer caused

Answer:

B.  a hearing by a federal court of appeals

If any state tribunal decides a federal question and the litigant has no further remedy within the state court system, a hearing by a federal court of appeals.

Oral argument in the court of appeals is a prearranged conversation between the judges and the appellate attorneys that focuses on the legal principles at issue. Each party has a limited amount of time usually around 15 minutes to make their case to the judge.

If a litigant loses in a federal court of appeals or in the highest court of a state, they may ask the supreme court to reconsider the case by filing a petition.

Only when a case concerns an extremely significant legal concept or when two or more federal appellate courts have disagreed on how to interpret a statute will the court normally consent to consider it. Additionally, there are a few rare instances where the Supreme Court is mandated by the law to consider an appeal.

Find more on federal court :

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Answer:

Object A runs into Object B. The force that object A has will transfer from Object A to Object B. But during this transfer you will loose some energy, therefore object B will not travel as fast.

Explanation:

Answer:

Plankton is the correct answer I think

Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related.  When we graph P vs V, the area under the line is a trapezoid.  So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³.  So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa.  So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

Answer:

4 x 10⁻⁴ J

Explanation:

C = 5000 pF, V = 400 V

Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J

Explanation:

n=75 kg

v=3m/s

m=n×v

m=75x3

m= 225n