9. P waves move faster than S waves A. True B. False

Explanation:

This question is not feasible. There is no way to calculate the energy needed because the question is missing the final temperature

The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:

Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.

          dV = - E . ds

          E = [tex]- \frac{dV}{ds} \ \hat s[/tex]  

Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.

Let's apply this expression for each section of the given graph:

1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.

  The derivative of a constant is zero.

        E = 0

2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.

We look for the equation of the line.

       V-V₀ = m (x- x₀)

We carry out the derivative.

      E = - m i ^

The slope (m) is:

       [tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]  

Let's calculate.

       [tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]  

Let's substitute.

       E =  [tex]2 \hat i \ V/m[/tex]  

3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.

We look for the equation of the line and we derive.

      E = - m i ^

Let's  substitute.

      [tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]  

    E = - 4 [tex]\hat i[/tex] V / m

4) From x₀ = 4.5 m to x_f = 6m.  The potential is constant and the derivative of a constant is zero.

      E = 0

5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]  

      E = - 0.5 [tex]\hat i[/tex] V/m

6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]

Let's substitute.

       E = 2 [tex]\hat i[/tex] V/m

7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]

Let's substitute.

       E = 1 [tex]\hat i[/tex]  V/m

In the attachment we can see these Electric fields as a function of distance.

In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:

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42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

[tex]x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;[/tex]

[tex]\Rightarrow mg\sin{\theta} = F[/tex]

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at [tex]\theta.[/tex] Solving for the angle, we get

[tex]\sin{\theta} = \dfrac{F}{mg}[/tex]

or

[tex]\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)[/tex]

[tex]\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right][/tex]

[tex]\;\;\;=42.9°[/tex]

Answer:

FA = 2FB

Force on spring A is twice the Force on spring B

Explanation:

F = kx

FB = (kB)x

FA = (kA)x

FA= (2kB)x

FA = 2(kB)x

FA = 2FB

The force [tex]F_A[/tex] needed to stretch spring A is going to be twice as much as the force [tex]F_B[/tex] needed to stretch spring B.

Explanation:

We know that the spring constants are related as

[tex]k_A = 2k_B[/tex]

The force [tex]F_A[/tex] needed to stretch spring A is given by

[tex]F_A = -k_Ax[/tex]

Also, the force [tex]F_B[/tex] needed to stretch spring is

[tex]F_B = -k_Bx[/tex]

Taking the ratio of the forces, we get

[tex]\dfrac{F_A}{F_B} = \dfrac{-k_Ax}{-k_Bx} = \dfrac{k_A}{k_B}[/tex]

Since [tex]k_A = 2k_B,[/tex] the equation above becomes

[tex]\dfrac{F_A}{F_B} = \dfrac{2k_B}{k_B} = 2[/tex]

or

[tex]F_A = 2F_B[/tex]

This shows that since the spring constant of spring A is twice as large as that of spring B, the force needed is going to be twice as large.

None of the given options is correct based on the relationship for the components of the crate's weight.

The given parameter:

The vertical component of the force on the crate is calculated as follows;

[tex]F_y = W \times cos(q)\\\\F_y = F_gcos(q)[/tex]

The horizontal component of the crate is calculated as follows;

[tex]F_x = F_g \times sin(q)\\\\F_x = F_g sin(q)[/tex]

Thus, we can conclude that none of the given options is correct based on the relationship for the components of the crate's weight.

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Answer:

Speed of light =m/s

wavelength = m

frequency = ?

we have

Speed = frequency × wavelength

[tex]3* 10^8[/tex] = frequency × [tex]5 * 10^{-7}[/tex]

 Frequency = [tex]\frac{3*10^8}{5*10^{-7}}=6*10^{14}[/tex]hz

Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

First of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f× λ

All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.

In this case, you know:

Replacing in the definition of propagation speed:

3×10⁸ m/s = f× 5×10⁻⁷ m

Solving:

3×10⁸ m/s ÷ 5×10⁻⁷ m= f

f= 6×10¹⁴ Hz

In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

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Answer:

goalindia

goalindiaGiven:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kg

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 W

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s- So, the velocity of the car will be 0.61❤

If it takes 4.4 Newton to lift 1 pound, then to lift 2 pounds the force required will be equal to 8.8 Newton.

A force in physics is an input that has the power to change an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe forces.

Being such a vector quantity, a force does have magnitude and direction. The SI unit metric newton is used to measure it (N). The letter F stands for force.

According to Newton's second law's original formulation, an object's net force is equal to the speed that its momentum is changing over time.

As per the given data in the question,

It takes 4.4 N to lift 1 pound,

Let the total force required to lift 2 pounds is x.

x = (4.4 × 2)/1

x = 8.8 N

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Answer:

hertz (Hz)

The number of periods or cycles per second is called frequency. The SI unit for frequency is the hertz (Hz).

Answer: the answer is hertz! if you’re on plato it’s option C. have a nice day! :)

Explanation:

Crazy Wally Ok Ok ok hhahahaha

The work done if the forklift is operating at full capacity is 11,900 J.

We have to recall that power is defined as the rate of doing work. The rate of doing work is defined as;

Power = Work done/time taken

When;

Power = 4950 W

Time taken = 2.40 s

Work done = Power × time taken

Work done = 4950 W × 2.40 s

Work done = 11,900 J

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Answer:

ghgivjgifigo ra together aigig disgust u hoodie

Answer:

I believe The Knowledge that we apply everyday or based on these methods and discoveries may represent us

Explanation:

Answer:

lithosphere

Explanation:

hope this helps you!!

Answer:

a. petrification

Explanation:

tar seeps = natural trade that, because of its close proximity to the ground surface, seeps from the cracks in the Earth or between rocks forming pits or pools (tar pits)

amber = fossilized resin produced by extinct coniferous trees, typically yellow in color

mummification = a process in which the skin and flesh of a corpse can be preserved by embalming and drying

Answer:

fifth harmonic

Explanation:

Explanation:

We use the Theorem of conservation of mechanical energy for finding the velocity when it strikes the ground:

Ei = Ef

Ki + Ui = Kf + Uf

Ui = Kf

m g h = 1/2 m v^2

v = sqrt(2gh)

So the momentum will be:

p = mv = m * sqrt(2gh)

Answer:

Decant it.

Explanation:

Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.

Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.

Answer:

D.

correct me if im wrong

brainlest plsss<333

Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   [tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}[/tex]

     Subsequently,

                                                  [tex]F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})[/tex]

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    [tex]v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}[/tex]

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 [tex]m \ s^{-1}[/tex].

                                                    [tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}[/tex]

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

Answer:

when the ball is at its highest point

Explanation:

Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.

Answer:

Explanation:

We can subtract directly the corresponding components and check using the parallelogram rule.

Explanation:

Have a look:

enter image source here

Where, graphically, I used the fact that:

→

A

−

→

B

=

→

A

+

(

−

→

B

)

For the magnitude we use Pythagoras (with the components) to get:

∣

∣

∣

→

A

−

→

B

∣

∣

∣

=

√

(

−

1

)

2

+

(

5

)

2

=

√

1

+

25

=

√

26

≈

5.1

For the direction I can see that will be

90

∘

from the

x

axis up to the

y

axis, plus the little bit passed the

y

axis given as:

θ

=

arctan

(

1

5

)

=

11.3

∘

giving in total: angle

=

90

∘

+

11.3

∘

=

101.3

∘

Answer:

force is simply a push or a pull in a particular direction. The unit is N (newton)

Explanation:

hope this helped

Answer:

In physics force is an influence which changes the motion of an object which has mass

Its unit is N or kgms/2

Answer:

Explanation:

a) F = ma

a = F/m

a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s

b) t = v/a

t = 200 / 7.2 x 10⁻⁶

t = 2.8 x 10⁷ s       about 10½ months

c) v² = u² + 2as

s = (v² - u²) / 2a

s = (200² - 0²) / (2( 7.2 x 10⁻⁶))

s = 2.8 x 10⁹ m    nearly 7 times around the earth

And all this assumes NO FRICTION.

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

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Answer:

D. High frequency and short wavelengths.

Explanation:

If a wave is high in energy it will have a higher frequency.

High frequency = short wavelengths

Answer:

a

Explanation:

you will find out that an object that has more weight requires more energy to move than the one with less weight

Answer:

The main condition necessary for a controlled thermonuclear fusion:

The ions must be held together in close proximity at high temperature with a confinement time long enough to avoid cooling.