Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m

Answer: R=24.2Ω

Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:

P=V.i

P=R.i²

[tex]P=\frac{V^{2}}{R}[/tex]

The resistance of the system is:

[tex]P=\frac{V^{2}}{R}[/tex]

[tex]R=\frac{V^{2}}{P}[/tex]

[tex]R=\frac{110^{2}}{500}[/tex]

R = 24.2Ω

For the device, resistance is 24.2Ω.

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Answer:

d. unchanged.

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

f is the frequency

λ is the wavelength of the sound wave.

Answer:

   θ  = 45º

Explanation:

The light that falls on the second polarized is polarized, therefore it is governed by the law of Maluz

              I = I₀ cos² θ

in the problem they ask us

            I = ½ I₀

let's look for the angles

             ½ I₀ = I₀ cos² θ

             cos θ  = √ ½ = 0.707

            θ  = cos 0.707

           θ  = 45º

Answer:

constructive interferencia  0, 1 , 2 m

destructive inteferencia   1/4, 3/4. 5/4, 7/4 m

Explanation:

This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression

            Δr / λ = Φ / 2π

            Δr = Φ/2π   λ

let's apply this expression to our case

λ = 1 m

            Δr = Φ 1 / 2π

We have constructive interference for angle of  Φ = 0, 2π, ...

let's find the values ​​where they occur

  Φ         Δr

   0          0

  2π         1

  4π        2

Destructive interference occurs by    Φ = π /2, 3π / 2, ...

 Φ          Δr

 π/2       ¼ m

 3π /2    ¾ m

5π /2     5/4 m

7π /2      7/4 m

Answer:

269 lb

Explanation:

We first find the tangential acceleration, a on the drums

a = Δv/Δt since the speed of the belt is 11 ft/s after 0.16 s, Δv = 11 ft/s and Δt = 0.16 s

a = Δv/Δt = 11 ft/s ÷ 0.16 s = 68.75 ft/s²

Since torque τ = Tk = Iα where I = moment of inertia of larger drum = Mk² where m = mass of larger drum = 4 lbs, k = radius of gyration of larger drum = 1.25 inches, T = tension due to larger drum and α = angular acceleration of larger drum.

So, T = Iα/k = Mk²α/k = Mαk = Ma (since a = αk )

T = 4 lbs × 68.75 ft/s² = 275 lb

The tension due to the smaller drum is T' = 6 lb  .

So the net tension in the belt is T'' = T - T' = 275 lb - 6 lb = 269 lb

Answer:

a) 0.56°C

b) 273.56 K

Explanation:

If we want to convert from Fahrenheit scale to Celcius scale we use the formula;

T(°C) = (T(°F) - 32) × 5/9

Where T(°F) = 33°F

Hence;

T(°C) = (33°F - 32) × 5/9

T°C = 0.56°C

b)

T(K) = T°C + 273

T(K) = 0.56 + 273

T(K) = 273.56 K

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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Answer:

Torque = 0.012 N.m

Explanation:

We are given;

Mass of wheel;m = 750 g = 0.75 kg

Radius of wheel;r = 25 cm = 0.25 m

Final angular velocity; ω_f = 0

Initial angular velocity; ω_i = 220 rpm

Time taken;t = 45 seconds

Converting 220 rpm to rad/s we have;

220 × 2π/60 = 22π/3 rad/s

Equation of rotational motion is;

ω_f = ω_i + αt

Where α is angular acceleration

Making α the subject, we have;

α = (ω_f - ω_i)/t

α = (0 - 22π/3)/45

α = -0.512 rad/s²

The formula for the Moment of inertia is given as;

I = ½mr²

I = (1/2) × 0.75 × 0.25²

I = 0.0234375 kg.m²

Formula for torque is;

Torque = Iα

For α, we will take the absolute value as the negative sign denotes decrease in acceleration.

Thus;

Torque = 0.0234375 × 0.512

Torque = 0.012 N.m

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Answer:

The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]

Explanation:

From the question we are told

    The diameter of the loop is  [tex]d = 6.7 cm = 0.067 \ m[/tex]

    The magnetic field is  [tex]B = 1.27 \ T[/tex]

    The time taken is  [tex]dt = 0.16 \ s[/tex]

Generally the induced emf is mathematically represented as

          [tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]

Where  N =  1 given that it is only a circular loop

            [tex]\Delta \phi = \Delta B * A[/tex]

Where  [tex]\Delta B = B_f - B_i[/tex]

   where [tex]B_i[/tex] is  1.27 T  given that the loop of wire was initially in the magnetic field

    and  [tex]B_f[/tex] is  0 T given that the loop was removed from the magnetic field

Now the area of the of the loop is evaluated as

          [tex]A = \pi r^2[/tex]

Where r is the radius which is mathematically represented as

       [tex]r = \frac{d}{2}[/tex]

substituting values

       [tex]r = \frac{0.067}{2}[/tex]

        [tex]r = 0.0335 \ m[/tex]

So

         [tex]A = 3.142 * (0.0335)^2[/tex]

          [tex]A = 0.00353 \ m^2[/tex]

So

      [tex]\Delta \phi = (0- 127)* (0.00353)[/tex]

      [tex]\Delta \phi = -0.00448 Weber[/tex]

=>    [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]

=>   [tex]\epsilon = 0.0280 \ V[/tex]

Answer:

Therefore maximum stretch is y2 = 32.36 m

Explanation:

In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium

        [tex]F_{e}[/tex] - W = 0

         k Δx = mg

         k = mg / Δx

         k = 80 9.8 / (30-20)

         k = 78.4 N / m

now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m

starting point. When will you jump

         Em₀ = U = mg y

final point. Just when the rope starts to stretch

         [tex]Em_{f}[/tex] = K = ½ m v²

         Em₀ = Em_{f}

          mg y = ½ m v²

          v = √ 2g y

          v = √ (2 9.8 20)

          v = 19.8 m / s

now all kinetic energy is transformed into elastic energy

starting point

            Em₀ = K = ½ m v²

final point

            Em_{f} = [tex]K_{e}[/tex] + U = ½ k y² + m g y

            Emo = Em_{f}

           ½ m v² = ½ k y² + mgy

            k y² + 2 m g y - m v² = 0

we substitute the values ​​and solve the quadratic equation

            78.4 y² + 2 80 9.8 y - 80 19.8² = 0

            78.4 y² + 1568 y - 31363.2 = 0

              y² + 20 y - 400 = 0

              y = [- 20 ±√ (20 2 +4 400)] / 2

              y = [-20 ± 44.72] / 2

the solutions are

              y₁ = 12.36 m

              y₂ = 32.36 m

These solutions correspond to the maximum stretch and its rebound.

Therefore maximum stretch is y2 = 32.36 m

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

Answer:

Capacitor, is the right answer.

Explanation:

The unknown element is a Capacitor.

Below is the calculation that proves that it is a capacitor.

We know that for the Capacitor

i = Imax × sin(wt+(pi/2)).

i = Imax × sin ((2 × pi/T) × (T/4) + (pi/2))

i = Imax × sin(3.142) = 0 A

at, t = T/2

wt = (2 × pi/T) × (T/2) = pi

wt + (pi/2) = pi + (pi/2) = ( 3 × pi/2) =

i = Imax × sin(3 × pi/2) = -Imax

Which is in a correct agreement with capacitor  therefore, the answer is a Capacitor.

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]

[tex]0.8 = e^{-\frac{88}{R} }[/tex]

[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]

[tex]ln(0.8) = -\frac{88}{R}[/tex]

[tex]R = -\frac{88}{ln(0.8)}[/tex]

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

Answer:

0.09N, attractive

Explanation:

It can be deducted from the question that the currents are arranged in parallel settings, then it is obvious that the force on each of the wire will be attractive toward the other wire.

the magnitude of force can be determined by using below formula;

F2 = (μ₀/2π)(I₁I₂/d)I₂

μ₀ = constant = 4π × 10^-7 H/m,

I₁, I₂ = currents= 30A

L = the length o the wire=30m

d = distance between these two wires= 0.06m

Since the current are arranged in the same direction, they exhibit attractive force on each other.

Then plugging the values Into the formula above we have

F₂ = (4π × 10^-7 T.m/A)/2π) × ((30A)²/ 0.06m)× 30 m

= 0.09 N, attractive

Therefore, the magnitude and direction of the force is 0.09 N, attractive

A) The acceleration of the electron along the x -axis is ; 4.57 * 10⁻¹¹ m /s²

B) The speed that would result in the electron experiencing an acceleration  along the x-axis is 4.57 * 10⁻¹¹  * time  m/s

Given Data :

Electric field ( E ) = ( 2.60i + 5.90j ) V/m

Magnetic field ( B ) = 0.400 k T

Velocity ( v ) = 8.0i m/s

Applying Lorentz force

F = q ( E + ( v * B ) )

  = 1.6 * 10⁻¹⁹ ( 2.60 i  +  5.90 j  + ( 8.0 i * 0.4 k ) ) N

  = 1.6 * 10⁻¹⁹ ( 2.60 i  +  5.90 j + ( 3.2 ( -j ) ) N

  = 1.6 * 10⁻¹⁹ (  2.60 i  + 2.70 j ) N

Ax = 4.57 * 10⁻¹¹ m /s²

Ax = Fx / Mc

    = ( 1.6 * 10⁻¹⁹ * 2.60 ) / 9.1 * 10⁻³¹

    = ( 4.16 * 10⁻¹⁹ ) / 9.1 * 10⁻³¹

    = 0.457 * 10¹²

    = 4.57 * 10⁻¹¹ m /s²

Therefore The speed that would result in the electron experiencing an acceleration  along the x-axis is 4.57 * 10⁻¹¹  * time

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Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

Answer:

a,)3.042s

b)4.173s

c)3.281s

Explanation:

For a some pendulum the period in seconds T can be calculated using below formula

T=2π√(L/G)

Where L = length of pendulum in meters

G = gravitational acceleration = 9.8 m/s²

Then we are told to calculate

(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?

Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then

use G = 9.8 + 3.0 = 12.8 m/s²

Period T=2π√(L/G)

T= 2π√(3/12.8)

T=3.042s

b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?

G = 9.8 – 3.0 = 6.8 m/s²

T= 2π√(3/6.8)

T=4.173s

C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?

Net acceleration is

g'= √(g² + a²)

=√(9² + 3²)

Then period is

T=2π√(3/11)

T=3.281s

Answer:

a) 3.6 ft

b) 12.4 ft

Explanation:

Distance between mirrors = 6.2 ft

difference from from the mirror you face = 1.8 ft

a) you stand 1.8 ft in front of the mirror you face.

According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,

your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft

b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.

the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,

the first image of your back will be 4.4 ft into the back mirror,

therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft

Answer:

  t = 1.81 min ,     the correct answer is c

Explanation:

This is a missile throwing exercise

The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation

             y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

the final height is y = 0 and the initial height is y₀ = 22000 m

            0 = y₀ + 0 - ½ g t²

            t = √y 2y₀ / g

let's calculate

           t = √(2  22000 / 3.72)

           t = 108.76 s

let's reduce to minutes

           t = 108.76 s (1 min / 60 s)

           t = 1.81 min

The correct answer is c

Answer:

A) 0.4 mA

B) 0.03 mA

Explanation:

Given that

voltage source, V = 120 V

to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.

mathematically, Ohms Law, V = IR

V = Voltage

I = Current

R = Resistance

from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have

120 = I * 300*10^3 Ω

making I the subject of the formula,

I = 120 / 300000

I = 0.0004 A

I = 0.4 mA

Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.

B, we have Resistance, R = 4000kΩ

Substituting like in part A, we have

120 = I * 4000*10^3 Ω

I = 120 / 4000000

I = 0.00003 A

I = 0.03 mA

This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA

The current through a person will be:

a) 0.4 mA

b) 0.03 mA

Given:

Voltage, V = 120 V

It states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.

Ohms Law, V = I*R

where,

V = Voltage

I = Current

R = Resistance

a)

Given: Resistance=  300kΩ

[tex]120 = I * 300*10^3 ohm\\\\I = 120 / 300000\\\\I = 0.0004 A[/tex]

Thus, current will be, I = 0.4 mA

b)

Given: R = 4000kΩ

[tex]120 = I * 4000*10^3 ohm\\\\I = 120 / 4000000\\\\I = 0.00003 A[/tex]

Thus, current will be, I = 0.03 mA

From calculations, we observe that nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA.

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Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Answer:

Explanation:

According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.

Mathematically the law is stated as

F= BIL

given data

Magnetic field B=  0.43T

Current I= 4.9 A

length of conductor L= 3.3cm to meter , 3.3/100=  0.033 m

Applying the formula the force is calculated as

F= 0.43*4.9* 0.033= 0.069 N

According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction

Answer:

a)1984.5nm

b)523mm

Explanation:

A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference

θ=(m+0.5)λ where m= 1,2.3....

Where given from the question the 4th dark Fringe which will take place at m= 3

θ=7/2y

Where y= 567nm

= 7/2(567)=1984.5nm

But

B)tan θ ≈ y/d

And sinθ = mλ/d

y=mλd when m= 1 which is the first bright we have

Then y=(1× 567.D)/d

But the distance from Central to the 4th dark Fringe is 1.83cm then

y= 7λD/2d= 1.83cm

D/d=(2)×(1.83×10^-2)/(7×567×10^-9)

=92221.5

y= (567×10^-9)× (92221.5)

=0.00523m

Therefore, the distance between the first and center is y1-y0= 523mm

Answer:gamma ray

Explanation:

Answer: the furthest from the end prying out the nail

Answer:

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the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

Explanation:

It depends on what stage of the mission you're talking about.

==>  While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.

==>  When the engines ignite, their thrust (d) is greater than the force of gravity.  So the net force on the rocket is upward, and the spacecraft accelerates upward.

==>  After the engines shut down, the net force acting on the rocket is due to Gravity (c).

. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.  

. . . If it has enough horizontal speed, it enters Earth orbit.  

. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.    

A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.

A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.

What's the net pressure of unbalanced?

If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.

Learn more about the thrust of the rocket engine. here:  https://brainly.com/question/10716695

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Answer:

v = 0.89 c = 2.67 x 10⁸ m/s

Explanation:

The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:

t = t₀/√[1 - v²/c²]

where,

t = time measured by the person on earth = 2.50 s

t₀ = rest time of the intergalactic rock star = 1.30 s

v = relative speed of the rock star = ?

Therefore,

2.5 s = (1.3 s)/√[1 - v²/c²]

√[1 - v²/c²] = 1.3/2.5

√[1 - v²/c²] = 0.52

[1 - v²/c²] = 0.52²

[1 - v²/c²] = 0.2074

v²/c² = 1 - 0.2074

v²/c² = 0.7926

v/c = √0.7926

v = 0.89 c

where,

c = speed of light = 3 x 10⁸ m/s

v = (0.89)(3 x 10⁸ m/s)

v = 0.89 c = 2.67 x 10⁸ m/s

The ball's horizontal position in the air is

[tex]x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t[/tex]

It hits the wall when [tex]x=17.5\,\mathrm m[/tex], which happens at

[tex]17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s[/tex]

Meanwhile, the ball's vertical position is

[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2[/tex]

where [tex]g[/tex] is the acceleration due to gravity, 9.80 m/s^2.

At the time the ball hits the wall, its vertical position (relative to its initial position) is

[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}[/tex]