If a jet travels 350 m/s, how far will it travel each second?

Assuming ideal conditions, Boyle's law says that

P₁ V₁ = P₂ V₂

where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.

So you have

(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂

==>   V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³

Answer:The radioactive waste

Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei

Answer:

Friction always opposes the motion

HAVE A GREAT DAY :)

(E) c

Explanation:

The speed of light is always equal to c regardless of the relative motion of the light source.

Answer:

Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.

Answer:

   t = 2.09 10⁻³ s

Explanation:

We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance

let's start with Newton's second law

        ∑ F = m a

the force is electric

        F = q E

we substitute

        q E = m a

        a = [tex]\frac{q}{m} \ E[/tex]

        a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]

        a = 1.37 10³ m / s²

now we can use kinematics

        x = v₀ t + ½ a t²

indicate that rest starts v₀ = 0

        x = 0 + ½ a t²

        t = [tex]\sqrt{\frac{2x}{a} }[/tex]

        t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]

        t = 2.09 10⁻³ s

The velocity of B after the collision is obtained as -2.6 m/s.

Now we now that the  principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.

Thus;

(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)

-7 = 10 + 6.5v

-7 - 10 = 6.5v

v = -7 - 10 /6.5

v = -2.6 m/s

Hence, the velocity of B after the collision is obtained as -2.6 m/s.

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Answer:

a) Dielectric constant ( λ ) = 750 * 10^-9 m

   minimum thickness of coating ( d )  = 187.5 nm

b) 3.6%

Explanation:

Given data:

wavelength of red light in air = 750 nm

εr = 4

μr = 1,  σ = 0

a) Determine the required dielectric constant and min thickness of coating used

Refractive index of coating ( n ) = √εr * μr =  √4*1 = 2

the refractive index of glass( ng)  = 1.5 which is < 2

λ = 750 * 10^-9 m

Dielectric constant ( λ ) = 750 * 10^-9 m

To determine the minimum thickness we will apply the formula below

d = m λ/2n ;  where  m = 1

∴ d = 750 nm / 2 ( 2 )

      = 187.5 nm

minimum thickness of coating ( d )  = 187.5 nm

b) Determine the percentage of the incident power that will be reflected

R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2

   = [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2

   = 0.03628 =  3.6%

Answer:

  Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

        a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

   a = 2 lines 1/600

   a = 2/600

    a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

      tan θ = y / L

as the measured angles are small

      tan θ = sin θ / cos θ sin θ

      sin θ = y / L

we substitute

     a  y/L = λ

     y = λ L / a

for λ = 400 10-9 m

      I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

      i = 346.89 10⁻⁶ m

f

or λ = 700 nm

        y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

        y_f = 609.609 10⁻⁶ m

the separation of this spectrum

        Δr = v_f - i

        Dr = (609.609 - 346)  10 ⁻⁶

        Dr = 263 10⁻⁶ m

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0  = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.  

So, option B is the answer

Answer:

Work Done= 3150J

Power= 1.75W

Explanation:

Work Done= Force x the distance travelled in the direction of the force (W= f x d)

Weight is a force, i think the qn. stated it wrongly, it should be 70N not 70kg.

Work Done= 70 x 45

=3150J

Power= Work Done/Time

=3150/(30x60)

*convert minutes to seconds since the S.I. unit of Power is joules/seconds(J/s) or watts(W)

=1.75W

Answer:

6000 lines/cm

Explanation:

From the question we are told that:

Grating 1=4000 lines/cm

Grating 2=6000 lines/cm

Generally The Spread of fringes is Larger when the Grating are closer to each other

Therefore

Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm

285.3 days

Explanation:

The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write

[tex]F_c = F_G[/tex]

or

[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]

where M is the mass of the star and R is the orbital radius around the star. We know that

[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]

where C is the orbital circumference and T is orbital period. We can then write

[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]

Isolating [tex]T^2[/tex], we get

[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]

Taking the square root of the expression above, we get

[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]

which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as

[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]

[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]

Answer:

a)  [tex]P=0.80[/tex]

b)  [tex]1.25Hz[/tex]

c)  [tex]A=25cm[/tex]

Explanation:

From the question we are told that:

Travel Time [tex]T=0.40s[/tex]

Distance [tex]d=50cm[/tex]

a)

Period

Time taken to complete one oscillation

Therefore

[tex]P=2*T\\\\P=2*0.40[/tex]

[tex]P=0.80[/tex]

b)

Frequency is

[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]

[tex]1.25Hz[/tex]

c)

Amplitude:the distance between the mean and extreme position

[tex]A=\frac{50}{2}[/tex]

[tex]A=25cm[/tex]

Answer:

[tex]I_2=6.8A[/tex]

Explanation:

From the question we are told that:

Turns of inner coil [tex]N_1=120[/tex]

Radius of inner coil [tex]r_1=0.012m[/tex]

Current of  inner coil [tex]I_1=6.0A[/tex]

Turns of Outer coil [tex]N_2=150[/tex]

Radius of Outer coil [tex]r_2=0.017m[/tex]

Generally the equation for Magnetic Field is mathematically given by

[tex]B =\frac{ \mu N I}{2R}[/tex]

Therefore

Condition for the net Magnetic field to be zero

[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]

[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]

[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]

[tex]I_2=6.8A[/tex]

∑ F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a

==>   a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²

To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².

The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:

Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)

F[tex]_{horizontal}[/tex] = 150 N × cos(60°)

F[tex]_{horizontal}[/tex] = 150 N × 0.5

F[tex]_{horizontal}[/tex] = 75 N

Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)

F[tex]_{vertical}[/tex] = 150 N × sin(60°)

F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)

F[tex]_{vertical}[/tex] ≈ 129.9 N

Now, let's calculate the net force in the horizontal direction:

Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]

F[tex]_{net horizontal}[/tex] = 75 N - 15 N

F[tex]_{net horizontal}[/tex] = 60 N

Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:

F[tex]_{net horizontal}[/tex] = m × a

60 N = 20 kg × a

Now, solve for acceleration (a):

a = 60 N / 20 kg

a = 3 m/s²

So, the acceleration of the box is 3 m/s².

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Answer:

杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音

Explanation:

莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd

Answer:

B velocity

Explanation:

Answer:

I = 15,645. kg*m/s or 15,645 N*s

Explanation:

I = m(^v)

I = 1050kg((26.2m/s-11.3m/s)

I = 15,645. kg*m/s

Answer:

    T₂ > T₁ > T₃ >T₄

Explanation:

In a simple harmonic motion the angular velocity is

         w = [tex]\sqrt{\frac{k}{m} }[/tex]

angular velocity and period are related

         w = 2π / T

we substitute

         T = [tex]2 \pi \ \sqrt{\frac{m}{k} }[/tex]

let's find the period for each case

a) mass m

   spring constant k

          T₁ = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) mass 2m

   spring constant k

          T₂ = 2π [tex]\sqrt{\frac{2m}{k} }[/tex]

          T₂ = T₁ √2

          T₂ = T₁ 1.41

c) masses 3m

   spring constant 6k

          T₃ = 2π [tex]\sqrt{\frac{3m}{6k} }[/tex]

          T₃ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.5}[/tex]

          T₃ = T₁ 0.707

d) mass m

    spring constant 4k

          T₄ = 2π [tex]\sqrt{ \frac{m}{4k} }[/tex]

          T₄ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.25}[/tex]

          T₄ = T₁ 0.5

now let's order the periods in decreasing order

           T₂ > T₁ > T₃ >T₄

Answer:

1.74×10⁻³ m

Explanation:

Applying,

ε = Stress/strain............. Equation 1

Where ε = Young's modulus

But,

Stress = F/A.............. Equation 2

Where F = Force, A = Area

Strain = e/L.............. Equation 3

e = extension, L = Length.

Substitute equation 2 and 3 into equation 1

ε = (F/A)/(e/L) = FL/eA............. Equation 4

From the question,

Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,

A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²

Substitute these values into equation 4

5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)

Solve for e

e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)

e = 82.65/4.74925×10⁴

e = 1.74×10⁻³ m

The question is incomplete, the complete question is;

Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect.

Part A If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be

K/2.

K.

2K.

greater than 2K.

Answer:

2K

Explanation:

Given that the kinetic energy of photo electrons is given by;

K= E -Wo

Where;

K = kinetic energy

E= energy of incident photon

Wo = work function

But;

E= hf

Wo = fo

h= Plank's constant

f= frequency of incident photon

fo= Threshold frequency

So:

K= hf - hfo

Where the frequency of incident light is doubled;

K= 2hf - hfo

Hence, maximum kinetic energy of the emitted electrons in this case will be 2K

Answer:

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

Explanation:

Concepts and reason

The concept required to solve this problem is Newton’s second law of motion.

Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.

Fundamentals

According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:

F = maF=ma

Here, m is mass and a is the acceleration.

(a)

Rearrange the equation F = maF=ma for a.

a = \frac{F}{m}a=  

m

F

​

Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=  

m

F

​

 .

\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}  

a=  

(2300kg+2400kg)

18,000N

​

=  

(4700kg)

18,000N

​

=3.83m/s  

2

​

(b)

Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 for a and 2400 kg for m in the equation F = maF=ma .

\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}  

F=(2400kg)(3.83m/s  

2

)

=9120N

​

Ans: Part a

The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 .

Part b

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².

The force of the SUV's bumper on the truck's bumper is 10000N

Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.

According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.

F = ma

And, a =  F/m

Given, the mass of the ruck , m = 2000 Kg

The mass of the SUV, M = 2500 Kg

The total mass of the both = 2000 + 2500 = 4500 Kg

The maximum force on the trick , F = 18000 N

The maximum acceleration of the truck can give the SUV:

[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]

a = 18000/4500

a = 4 m/s²

The force of the SUV's bumper on the truck's bumper will be:

[tex]F_{max} -f= ma_{max}[/tex]

[tex]f= 18000-2000\times 4[/tex]

[tex]f =10000N[/tex]

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Explanation:

Given:

[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]

[tex]V = 12 V[/tex]

[tex]I = 1.2 A[/tex]

Recall that power P is given by

[tex]P = VI[/tex]

so the amount of energy dissipated [tex]\Delta E[/tex] is given by

[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]

[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

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

Where:

[tex]x_{o}[/tex] - Initial x-position, in meters.

[tex]v_{o,x}[/tex] - Initial x-velocity, in meters per second.

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

[tex]a_{x}[/tex] - x-acceleration, in meters per second.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:

[tex]x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}[/tex]

[tex]x(t) = 0.324\,m[/tex]

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

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

Where:

[tex]y_{o}[/tex] - Initial y-position, in meters.

[tex]v_{o,y}[/tex] - Initial y-velocity, in meters per second.

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

[tex]a_{y}[/tex] - y-acceleration, in meters per second.

If we know that [tex]y_{o} = 0\,m[/tex], [tex]v_{o,y} = -3.6\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{y} = 0\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:

[tex]y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}[/tex]

[tex]y(t) = -2.16\,m[/tex]

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ([tex]v_{x}[/tex]), in meters per second, is calculated by this kinematic formula:

[tex]v_{x}(t) = v_{o,x} + a_{x}\cdot t[/tex] (3)

If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-velocity of the skateboarder is:

[tex]v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)[/tex]

[tex]v_{x}(t) = 1.08\,\frac{m}{s}[/tex]

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

[tex]v_{y} = -3.6\,\frac{m}{s}[/tex]

The y-velocity of the skateboard is -3.6 meters per second.

Answer:

One can write F = K d^-2  where K = G M m

So dF/dd = -2 K d^-3 =   -2 K / d^3    (As d increases F decreases - it is opposite to the direction of F)

Answer:

1 ohm

Explanation:

First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:

1 / Eq = (1 / r₁) + (1 / r₂)

The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:

Eq = r₁ + r₂

Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:

1/E₁ = 1/2 + 1/2

1/E₁ = 1

E₁ = 1Ω

1/E₂ = 1/2 + 1/2

1/E₂ = 1

E₂ = 1Ω

This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:

E₃ = E₁ + E₂ = 1 + 1 = 2Ω

The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:

1/Eq = 1/E₃ + 1/2

1/Eq = 1/2 + 1/2

1/Eq = 1

Eq = 1Ω

Answer:

acoustic energy and mechanical energy

Explanation:

each type of sounds has to be tackled in their own way.