A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.

What is the electric potential difference through which the proton moved?

2.5 × 105 V
3.1 × 105 V
5.6 × 105 V
8.1 × 105 V

Answer:

a). Single replacement.

Explanation:

Because one element replaces another element in a compound

Answer:

isopropyl alcohol would form smaller droplets, because it has lower surface tension than water has

Explanation:

Ap3x

The droplets made of isopropyl alcohol instead of water be smaller due to surface tension.

The single drop of a liquid in the form of sphere is called droplet.

Water can form large dewdrops in nature. Isopropyl alcohol would form smaller droplets, because it has lower surface tension than water.

Surface tension is the property of the liquid to acquire minimum surface area.

Thus, droplets made of isopropyl alcohol instead of water be smaller.

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Answer:

[tex]4.47\ \text{km/h}[/tex]

Explanation:

[tex]\dfrac{da}{dt}[/tex] = Rate at which the distance between A and starting point of B is changing = -20 km/h

[tex]\dfrac{db}{dt}[/tex] = Rate at which the distance of B is changing = 15 km/h

[tex]\dfrac{dc}{dt}[/tex] = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = [tex]20\times 4=80\ \text{km}[/tex]

a = Distance remaining to the start point of B = [tex]110-80=30\ \text{km}[/tex]

b = Distance covered by B in 4 hours = [tex]15\times 4=60\ \text{km}[/tex]

Distance between A and B after 4 hours

[tex]c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}[/tex]

[tex]c^2=a^2+b^2[/tex]

Differentiating with respect to time we get

[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}[/tex]

The rate at which the distance between the ships is changing at 4 PM is [tex]4.47\ \text{km/h}[/tex].

The force 1 is tension force.

To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.

Thus, we can conclude that the correct option is (B).

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Answer: 5 km per hour

Explanation:

if in 10 km there is 2 hours, then 10 divided by 2 is 5.

Answer:

refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium.

Answer:

vcyl / vsph = 1.05

Explanation:

       [tex]K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)[/tex]

       [tex]K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)[/tex]

       [tex]v = \omega * R (3)[/tex]

       [tex]K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)[/tex]

       [tex]K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)[/tex]

        [tex]I_{sph} = \frac{2}{3} * M* R^{2} (7)[/tex]  

       [tex]K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)[/tex]

      [tex]K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)[/tex]

       [tex]\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)[/tex]

Answer:

Bb

Explanation:

Answer:

[tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]

[tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex]

Explanation:

[tex]\lambda[/tex] = Wavelength

[tex]\theta[/tex] = Angle

m = Order

Distance between grating is given by

[tex]d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}[/tex]

[tex]\lambda=656\ \text{nm}[/tex]

We have the relation

[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]

m = 1

[tex]\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}[/tex]

The first and second order angular deflection is [tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]

[tex]\lambda=410\ \text{nm}[/tex]

m = 1

[tex]\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}[/tex]

The first and second order angular deflection is [tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex].

Answer:

harmful effects

1. that will cause air pollution

2. that will destroy our earth

Answer:

useful effects of volcano are :-

harmful effects of volcano are:-

hope it is helpful to you ☺️

When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.

T

Answer:

the velocity is a second final to initial velocity of 39

Answer:

1.48×10⁻⁷ Newtons

Explanation:

From the question,

According to newton's law of universal gravitation.

F = Gmm'/r²........................ Equation 1

F = gravitational force, G = gravitational constant, m = mass of the first ball, m' = mass of the second ball, r = distance between the balls.

Given: m = m' = 8 kg, r = 17 cm = 0.17 m,

Constant : G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

F =  (6.67×10⁻¹¹×8×8)/(0.17²)

F = 1.48×10⁻⁷ N

Answer:

 K_f = 1881.6 J

Explanation:

To solve this exercise, let's start by finding the velocities of the bodies.

We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved

initial instant. Before the explosion

        p₀ = M v₀

final instant. After the explosion

        p_f = m₁ v + m₂ 0

the moeoto is preserved

         p₀ = p_f

         M v₀ = m₁ v

         v = [tex]\frac{m_1}{M}[/tex]  v₀

in the exercise they indicate that the most massive part has twice the other part

         M = m₁ + m₂

         M = 2m₂ + m₂ = 3 m₂

         m₂ = M / 3

so the most massive part is worth

        m₁ = 2 M / 3

we substitute

        v = ⅔ v₀

with the speed of each element we can look for the kinetic energy

initial

         K₀ = ½ M v₀²

Final

         K_f = ½ m₁ v² + 0

         K_f = ½ (⅔ M) (⅔ v₀)²

         K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)

         K_f = [tex]\frac{8}{27}[/tex]  K₀

the energy added to the system is

         ΔK = Kf -K₀

         ΔK = (8/27 - 1) K₀

         ΔK = -0.7 K₀

         K_f = K₀ + ΔK

         K_f = K₀ (1 -0.7)

         K_f = 0.3 K₀

let's calculate

         K_f = 0.3 (½ 64 14²)

         K_f = 1881.6 J

Answer:

0.54 sec

Explanation:

Answer:

Explanation:

Speed = 14.2 m/s

Distance = 514 m

Time = Distance / Speed

Time = 514 / 14.2

Time = 36.19 seconds

Answer:

The force will be "9.8 N".

Explanation:

The given values are:

mass,

m = 0.7 kg

M = 2

g = 9.8

Now,

⇒  [tex]\tau = T \alpha[/tex]

then,

⇒  [tex]\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R[/tex]

⇒  [tex]\frac{1}{2}m \ a_t=m(g-a_t)[/tex]

⇒  [tex]a_t=\frac{2g}{(\frac{m}{M} +2)}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2\times 9.8}{\frac{0.7}{2} +2}[/tex]

⇒      [tex]=8.34 \ m/s[/tex]

hence,

⇒  [tex]T=mg+M(g-a_t)[/tex]

On substituting the values, we get

⇒      [tex]=0.7\times 9.8+2(9.8-8.34)[/tex]

⇒      [tex]=6.86+2(1.46)[/tex]

⇒      [tex]=6.86+2.92[/tex]

⇒      [tex]=9.8 \ N[/tex]

Answer:

the net work done on the object is 80 J.

Explanation:

Given;

force applied on the object, F = 100 N

mass of the object, m = 20 kg

speed of the object, v = 0.2 m/s

distance moved by the object, d = 0.8 m

The net work done on the object is calculated as follows;

W = F x d

W = 100 N x 0.8 m

W = 80 J

Therefore, the net work done on the object is 80 J.

Answer:

239 rpm

Explanation: So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (or wheel revolutions) fit inside the 75,000 cm? In other words, if I were to peel this wheel's tread from the cart and lay it out flat, it would measure a distance of 100π cm. How many of these lengths fit into the entire distance covered in one minute? To find out how many of (this) fit into so many of (that), I must divide (that) by (this), so:

100πcm/rev

75,000cm/min

​750 min rev≈238.7324146RPM

Answer:

r₂ = 4 r

Explanation:

For this exercise let's use Newton's second law with the magnetic force

          F = q v x B

bold letters indicate vectors, the magnitude of this expression is

          F = q v B sin θ  

in this case we assume that the angle is 90º between the speed and the magnetic field.

If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal

           a = v² / r

let's use Newton's second law

           F = ma

           q v B = m v² / r

           r = [tex]\frac{qB}{mv}[/tex]

Let's apply this expression to our case.

Proton 1

             r = \frac{qB_1}{mv_1}

Proton 2

             r₂ = [tex]\frac{q \ B_2}{m \ v_2}[/tex]

in the exercise indicate some relationships between the two protons

*    v₁ = 2 v₂

    v₂ = v₁ / 2

*   B₂ = 2B₁

we substitute

           r₂ = [tex]\frac{q \ 2B_1}{m \ \frac{v_1}{2} }[/tex]

           r₂ = 4 [tex]\frac{qB_1}{mv_1}[/tex]

           r₂ = 4 r

B, the body at rest becomes reluctant to start moving or a body in motion becomes

reluctant and stop moving once in motion in a straight line

Answer:

12 min

Explanation:every 4 minutes is 1 mile

Answer:

The speed of the bird is 1.00% of the speed of sound.

Explanation:

The speed of the bird can be found by using the Doppler equation:

[tex] f = f_{0}(\frac{v - v_{r}}{v - v_{s}}) [/tex]

Where:

v: is the speed of sound = 343 m/s

f₀: is the frequency emitted = 1490 Hz

f: is the frequency observed = 1505 Hz

[tex]v_{r}[/tex]: is the speed of the receiver = 0 (it is stationary)

[tex]v_{s}[/tex]: is the speed of the source =?

The minus sign of [tex]v_{s}[/tex] is because the source is moving towards the receiver.

By solving the above equation for [tex]v_{s}[/tex] we have:

[tex] v_{s} = v - \frac{f_{0}*v}{f} = 343 - \frac{1490*343}{1505} = 3.42 m/s [/tex]

The above speed in terms of the speed of sound is:

[tex]\% v_{s} = \frac{3.42}{343}\times 100 = 1.00 \%[/tex]

Therefore, the speed of the bird is 1.00% of the speed of sound.  

I hope it helps you!

1: Ming

2: Chloe

3: 40m

Answer:

v2(final)=5 m/s

Explanation:

we are going to use the conservation of momentum here

m1*v1(initial)+m2*v2(initial)=m1*v1(final)+m2v2(final)

m1=7 kg  v1(initial)=10 m/s

m2=14 kg  v2(initial)=0 m/s (bc initially it is at rest)

v1(final)= 0 m/s (m1 stops moving after the collision)

v2(final)=?

7*10+14*0=7*0+14*v2(final)

70=14v2(final)

v2(final)=70/14 m/s=5 m/s

Answera black cube or dark colors cause dark colors suck in heat