19. Using the formula V - IX R, find the voltage in an electrical component with a current of 2 A, and a resistance of 5 omega

O A 20 v
O 8.5v
O C. 2.5 V
O D. 10 v

Answer:

d. the star is a member and also a part of an eclipsing binary star system.

Explanation:

If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).

The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.

Thus, option D is correct.

Answer:

soluble soluble soluble soluble

Explanation:

solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci

Answer:

a)   y = 3.73 m,  b)  t = 1.74 s,  c)  R = 40.99 m,

d) vₓ  = 23.49  m/s,   v_y = -8.5 m / s

Explanation:

This is a projectile launching exercise, we start by breaking down the initial velocity

          sin θ = v_{oy} / v₀

          cos θ = v₀ₓ / v₀

          v_{oy} = v₀ sin θ

          v₀ₓ = v₀ cos θ

          v_{oy} = 25 sin 20 = 8.55 m / s

          v₀ₓ = 25 cos 20 = 23.49 m / s

a) when the egg reaches the maximum height its vertical speed is zero

          v_y² = v_{oy}² - 2 g y

          0 = v_[oy}² - 2g y

           y = v_{oy}² / 2g

          y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]

          y = 3.73 m

b) flight time

          y = v_{oy} t - ½ g t²

the time of flight occurs when the body reaches the ground y = 0

          0 = (v_{oy} - ½ g t) t

The results are

          t₁ = 0s        this time is for using the body star

          v_{oy} - ½ g t = 0

           t = [tex]\frac{2v_{oy}^2}{g}[/tex]

           t = 2 8.55 / 9.8

           t = 1.74 s

c) the range

           R = v₀² sin 2θ / g

           R = 25² sin (2 20) / 9.8

           R = 40.99 m

d) speed at the point of arrival

horizontal speed is constant

           vₓ = v₀ₓ = 23.49  m/s

vertical speed is

           v_y = Iv_{oy} - g t

           v_y = 8.55 - 9.8  1.74

           v_y = -8.5 m / s

Answer:

The correct answer is option (A) that is KEA > KEB .

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

[tex]W=F_d- F_f_r_id-F_gh[/tex]

[tex]W=F_d-\mu_kmgdcos\theta-mgdsin\theta[/tex]

The change in kinetic energy is ,

   [tex]\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2[/tex]

At the top of the inclined plane , the velocity is zero

So,

[tex]\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2[/tex]

[tex]\Delta KE=-\frac{1}{2}m\nu_0^2[/tex]

From the work energy theorem , we have [tex]W=-\Delta K[/tex] in case of friction , so

[tex]\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

[tex]KE=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

For object A-

[tex]KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

For object B

[tex]KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta[/tex]

[tex]KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)[/tex]

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

[tex]KE_A >KE_B[/tex]

Therefore , option A is correct .

Answer:

300

Explanation:

the momentum is 300

p=mv

p=5×60

5×60 =300

Answer:

7374.4

Explanation:

I took the test

(filler so I can post)

The term used to describe the block of organs that are removed during the autopsy is Thoracic organs.

The organs that are removed during autopsy include:

The thoracic cavity contains organs and tissues that function in the respirator, cardiovascular, nervous and digestive system.

These thoracic organs include the following;

Thus, we can conclude that the term used to describe the block of organs that are removed during the autopsy is Thoracic organs.

Learn more here:https://brainly.com/question/25087653

Answer:

the required or need power is 115960.57 Watts  

Explanation:

First of all, we take down the data we can find from the question, to make it easier when substituting values into formulas.

mass of swordfish m = 650 kg

Cross - sectional Area A = 0.92 m²

drag coefficient C[tex]_D[/tex] = 0.0091

speed v = 30 m/s

density p = 1026 kg/m³

Now, we determine our Drag force F[tex]_D[/tex]

Drag force F[tex]_D[/tex]  = [tex]\frac{1}{2}[/tex] × C[tex]_D[/tex] × A × p × v²

Next, we substitute the values we have taken down, into the formula.

Drag force F[tex]_D[/tex]  = [tex]\frac{1}{2}[/tex] × 0.0091 × 0.92 × 1026 × (30)²

Drag force F[tex]_D[/tex]  = 4.294836 × 900

Drag force F[tex]_D[/tex]  = 3865.3524

Now, we determine the power needed P[tex]_w[/tex]

P[tex]_w[/tex] = F[tex]_D[/tex] × v

we substitute  

P[tex]_w[/tex] = 3865.3524 × 30

P[tex]_w[/tex] = 115960.57 Watts  

Therefore, the required or need power is 115960.57 Watts  

Answer:

I think D. Infrared radiation.

Answer:

infrared radition

Explanation:

valid

Answer:

charge = electrons + protons

=92+92

=184

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:

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

Answer:

Half life is 3.23 hours

Explanation:

Given

Decay rate at starting = 1160 decays per minute

Decay rate after 4 hours = 170 decays per minute

As we know know

[tex]N = N_0 *e ^{\Lambda *T}[/tex]

Substituting the given values, we get -

[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]

Also

[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]

Substituting the given values we get -

[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours

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:

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

Answer: D. An AC circuit

Explanation:

I took it on a test and it was correct ; )

Answer:

Earth has a much greater mass than objects on its surface

Answer:

assesses C.) muscular endurance

Explanation:

Answer:

Wind blows__wind energy ____causing the turbine to turn, rotating shaft works pump__mechanical___energy.

Motion of water moving up from well___kinetic energy___energy.

Water in tank which is positioned 5 feet above the ground level______potential energy

Explanation:

Wind has in it wind energy which is then used to rotate the turbine shaft which is a form of mechanical work and hence possess mechanical energy

Moving water posses kinetic energy and when this water is kept at some height it possess potential energy

Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!  

bit.[tex]^{}[/tex]ly/3a8Nt8n

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

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

I hope this helped!+*

Answer:

Nuclear fusion

Explanation:

Answer:

A

Explanation:

Answer:

Explanation:

The classification will be made into 3 categories, which are

Ones that shortens wavelengths

Ones that lengthens wavelengths

Ones that has no effect on wavelengths

Shortens wavelengths -> Increase frequency

Lengthens wavelengths -> Decrease frequency

No effect -> Increase amplitude, decrease amplitude, increase damping, decrease damping.

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