If acceleration is zero what statement about velocity is
true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing​

Answer:

[tex]4.12\times 10^{-5}\ J[/tex].

Explanation:

Given that,

Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]

Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

[tex]E=\dfrac{Q^2}{2C}[/tex]

Put all the values,

[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]

So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].

Answer: I think B.)

Explanation:

Answer:

W' = 1.66 x 10¹⁴ N

Explanation:

First, we will calculate the mass:

[tex]W = mg[/tex]

where,

W = weight on earth = 690 N

m = mass = ?

g = acceleration due to gravity on earth = 9.8 m/s²

Therefore,

[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]

Now, we will calculate the value of g on the neutron star:

[tex]g' = \frac{GM}{R^2}[/tex]

where,

g' = acceleration due to gravity on the surface of the neutron star = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of the Neutron Star = 1.99 x 10³⁰ kg

R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m

Therefore,

[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]

Therefore, the weight on the surface of the neutron star will be:

[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]

W' = 1.66 x 10¹⁴ N

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

Determine the Thrust developed

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D  ) = 2.6 m

first step : calculate the area of the duct

A = π/4 * D^2

   =  π/4 * ( 2.6)^2  = 5.3092 m^2

next : calculate the velocity of propeller

Punit = (A*v*β ) / 2  * V^2     ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

         = ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

Finally determine the thrust developed

F = Punit / V2

  = (1900 * 10^3) / ( 8.9480)

  = 212.3373 kN

Answer:

A force is a push or pull upon an object resulting from the object's interaction with another object.

Momentum is force or speed of movement.

Velocity defines the path of the motion of the frame or the object

Answer:

[tex]523269.9\ \text{N/m}[/tex]

Explanation:

q = Charge

r = Distance

[tex]q_1=25\ \text{C}[/tex]

[tex]r_1=3000\ \text{m}[/tex]

[tex]q_2=40\ \text{C}[/tex]

[tex]r_2=850\ \text{m}[/tex]

The electric field is given by

[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]

The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]

Answer:

T = 0.71 seconds

Explanation:

Given data:

mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.

We have to calculate time period when this same spring-mass system oscillates vertically.

As we know

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

This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating  vertically too remains the same.

Therefore, T = 0.71 seconds

Question: Two people stand facing each other at a roller-skating rink then push off each other. If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity of the boy skater if his mass is 50 kg?

Answer:

3 m/s

Explanation:

Applying,

The Law of conservation of momentum

Momentum of the girl skater = momentum of the boy skater

MV = mv...................... Equation 1

Where M = mass of the girl skater, V = velocity of the girl skater, m = mass of the boy skater, v = velocity of the boy skater

From the question, we were asked to calculate v

v = MV/m.................. Equation 1

Given: M = 30 kg, V = 5 m/s, m = 50 kg

Substitute these values into equation 1

v = (30×5)/50

v = 3 m/s

Hence the velocity of the the boy skater is 3m/s

Answer:

i dont know but i should know try g o o g l e

Explanation:

Answer:

4 seconds

Explanation:

Assuming that the object started from rest,

v = at

--> t = v/a = (12 m/s) / (3 m/s^2)

= 4 seconds

Answer:

number 1

Explanation:

they have common ancestors

Answer:

50j

Explanation:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Answer:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Explanation:

Answer:

You get Day and Night

It takes 24 hour

Answer:

Explanation:

The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

         Wₓ = 1200 sin 30 = 600 N

          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

        P + fr -Wx = 0

       fr = μ N

       fr = 0.20 1039.23

        fr = 207.85 N

we substitute

        P =  Wₓ -fr

        P = 600 - 207,846

        P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

         fr = 155.88 N

         P = 600 - 155.88

         P = 444.12 N

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

Answer:

Earth has a much greater mass than objects on its surface

Answer:

Nuclear fusion

Explanation:

Answer:

Truck [tex]\dfrac{g}{10}[/tex]

Road [tex]-\dfrac{g}{10}[/tex]

Explanation:

[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]

[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]

Frictional force is given by

[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]

Net acceleration is given by

[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]

The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.

Answer:

[tex]10.54\ \text{rad/s}[/tex]

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]

[tex]\omega_f[/tex] = Final angular velocity

t = Time = 34 s

[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]

[tex]\alpha[/tex] = Angulr acceleration

From the kinematic equations of angular motion we have

[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]

The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].

Answer:

Frictional Force

Explanation:

i think so

Answer:

  K_{total} = 8.91 J

Explanation:

In this exercise you are asked to find the kinetic energy of the can of coca-cola

         K_total = K_ {Translation} + K_ {rotation}

the translational kinetic energy is

         K_ {translation} = ½ m v²

the kinetic energy of rotation is

         K_ {rotation} = ½ I w²

The moment of inertia of a cylinder is

           I = ½ m r²

we substitute

          K_ {total} = ½ m v² + ½ (½ m r²) w²

angular and linear velocity are related

          v = w r

we substitute

          K_ {total} = ½ m v² + ¼ m r² v² / r²

          K_ {total} = m v² (½ + ¼)

          K_ {total} = ¾ m v²

let's calculate

          K_ {total} = ¾ 0.33 6.00²

           K_{total} = 8.91 J

Answer:

24 J

Explanation:

[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]

Answer:

trueeeeeeee..........mmmm...........

Answer:

Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Explanation:

If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.

Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Answer:

a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,

e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

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

a) find the period

angular velocity, frequency, and period are related

         w = 2π f = 2π / T

          f = 1 / T

          T = 1 / f

           T = 1 / 0.59

           T = 1.69 s

b) the spring constant

         w = 2π f

         w = 2π 0.59

         w = 3.70 rad / s

         w² = k / m

          k = w² m

          k = 3.70² 0.060

          k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

          x = A cos (wt + Ф)

speed is defined by

         v =[tex]\frac{dx}{dt}[/tex]

          v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

          v = A w

          v = 0.394 3.70

          v = 1.46 feet/s

d) maximum acceleration

            a = [tex]\frac{dv}{dt}[/tex]

            a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

            a = A w²

            a = 0.394 3.70²

            a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

            x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

           v = -A w sin (0 + Фi)

           0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

            x = A cos wt

            x = 0.394 cos 3.70t

we substitute

           0.1969 = 0.394 cos 370t

           3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

           3.70, t = 1.047

           t = 1.047 / 3.70

           t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

           v = - Aw sin wt

           v = - 0.394 3.70 sin 3.70 0.2826

           v = - 1,319 ft / s

           a = - A w² cos wt

           a = - 0.394 3.70² cos 3.70 0.2826

           a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

           K = ½ m v²

let's slow down to the SI system

           v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

           K = ½ 0.060 0.402²

           K = 4.8 10⁻³ J

           U = ½ k x²

           U = ½ 0.825 0.06²

           U = 1.49 10⁻³ J

Answer:

It will go up to 93.75 m before it is moving at 20 m/s

Explanation:

As we know that

[tex]v^2 - u^2 = 2aS[/tex]

here v is the final speed i.e 20 m/s

u is the initial speed i.e 5 m/s

a is the acceleration due to gravity i.e 2 m/s^2

Substituting the given values in above equation, we get -

[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters

A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.

An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.

By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.

The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.

To know more about Planet:

https://brainly.com/question/14581221

#SPJ2

Answer:

 θ = 32.4º

Explanation:

For this exercise let's use Malus's law

         I = Io cos² θ

in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical

         I₀ = 370/2 = 185 W / m²

this is the radiation that affects the second polarizer, let's apply the expression of Maluz

         θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])

         θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])

         θ = cos⁻¹ (0.844697)

         θ = 32.4º