What is the weight of a 3000g girl who is swimming in the Pacific Ocean?

Answer:

b) less friction is created

Answer:

B

Explanation:

Answer:

It is the physics that explains how everything works, the nature of the particles that make up matter and the forces with which they interact.

Its types: Electromagnetism, the strong nuclear force, and the weak nuclear force.

Hope this help :)

Answer:

During the process of photosynthesis, cells use carbon dioxide and energy from the Sun to make sugar molecules and oxygen. These sugar molecules are the basis for more complex molecules made by the photosynthetic cell, such as glucose.

Explanation:

yes.

Answer:

Sunlight, Carbon Dioxide, and Water

Explanation:

Technically minerals are in there too but when I learned this it was just Sunlight, Carbon Dioxide, and Water

The converging lens is also called a concave lens. The height of the image formed by the lens is 2.55 cm.

Using the lens formula;

1/f = 1/u + 1/v

f = focal length of the lens

u = object distance

v = image distance

Note that the focal length of a converging lens is positive

Substituting values;

1/12 = 1/28 + 1/v

1/v = 1/12 -  1/28

v = 8.4 cm

Magnification= image height/object height = image distance/object distance

image height = ?

object height = 8.5 cm

image distance = 8.4 cm

object distance =  28 cm

So

image height/8.5 = 8.4/28

image height = 8.5 × 8.4/28

image height = 2.55 cm

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Answer:

4N

Explanation:

a = (7-3)/5 = 0.8m/s^2

F = ma = (5)(0.8) = 4 Newtons

Answer:

the force between the building and the ball is non-conservative (friction-type force)

Explanation

Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.

When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.

When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall

Consequently, the force between the building and the ball is non-conservative (friction-type force

Even without wind the sun help evaporate the water so it would dry faster. The other towel is in a bag so the moist and water has nowhere to go, therefor staying in the towel.

Answer:

Current flow I =  20 ampere

Explanation:

Given:

Resistance R = 11 ohms

Voltage V = 220 volts

Find:

Current flow I

Computation:

Current flow I = V / R

Current flow I = 220 / 11

Current flow I =  20 ampere

The amount of current flow through the element is of 20 A.

Given data:

The magnitude of resistance of Electric stove is, R = 11 ohms.

The magnitude of potential difference in a line is, V' = 220 V.

Here we can simple go for Ohm's law. As per the Ohm's law, the potential difference across the element is proportional to the current flow and the resistance of the element.

The expression is,

V' = I × R

here, I is the amount of current flowing through the element.

Solving as,

220 = I × 11

I = 20 A

Thus, we can conclude that the amount of current flow through the element is of 20 A.

Learn more about the Ohm's law here:

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Answer:

momentum=mass×velocity

momentum =400kg×20m/s=8000kg.m/s

Answer:

0.01 m

Explanation:

Since the speed of light is 3.0×10^8 m/s

Use the equation,

Wavelength = speed ÷ frequency

Wavelength = 3.0×10^8 ÷ 3×10^10

Wavelength = 0.01m

Answer:

A white dwarf, also called a degenerate dwarf

Explanation:

sorry if im wrong im kind of du-m

Complete question is;

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.

The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m

where, for example, D_ad denotes the distance in meters between conductors a and d.

a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.

b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A

Answer:

A) M = 1.01 × 10^(-4) H/km

B) v_cd = 5.712 V/km

Explanation:

A) From the distances given in the question, we can deduce that;

D_ac = √(((2.5/2) - (1/2))² + 1.8²)

D_ac = 1.95 m

Also;

D_ad = √(((2.5/2) + (1/2))² + 1.8²)

D_ad = 2.51 m

I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;

φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))

Mutual inductance per km is given as;

M = φ_cd/I_a

Thus;

M = 4 x 10^(-7)( ln (2.51/1.95))

M = 1.01 × 10^(-7) H/m

Per km;

M = 1.01 × 10^(-7) × 1000

M = 1.01 × 10^(-4) H/km

B) voltage per km is gotten by;

v_cd = ωMI

Now, ω = 2πf = 2π × 60 = 377 rad/s

Thus;

v_cd = 377 × 1.01 × 10^(-4) × 150

v_cd = 5.712 V/km

Answer: J

Explanation:

Jason pushes on the top of the scales and Kerry pushes on the base. They hold the scales still. Jason's push measures 80N. What is the size of Kerry's push?

he pushes zero

Answer:

 L = 116.6 m

Explanation:

For this exercise we approximate the tunnel as a tube with one end open and the other closed, at the open end there is a belly and at the closed end a node, therefore the resonances occur at

               λ = 4L         1st harmonic

               λ = 4L / 3    third harmonic

               λ = 4L / 5    fifth harmonic

General term

               λ = 4L / n      n = 1, 3, 5,...    odd

                                    n = (2n + 1)      n are all integers

They indicate that two consecutive resonant frequencies were found, the speed of the wave is related to the wavelength and its frequency

            v = λ f

            λ = v / f

we substitute

            [tex]\frac{v}{f} = \frac{4L}{n}[/tex]

             L = [tex]n \frac{ v}{4f}[/tex]

for the first resonance n = n

             L = (2n + 1) [tex]\frac{v}{4f_1}[/tex]

for the second resonance n = n + 1

             L = (2n + 3) [tex]\frac{v}{4f_2}[/tex]

we have two equations with two unknowns, let's solve by equating

             (2n + 1) \frac{v}{4f_1}= (2n + 3) \frac{v}{4f_2}

             (2n + 1) f₂ = (2n +3) f₁

              2n + 1 = (2n + 3) [tex]\frac{f_1}{f_2}[/tex]

              2n (1 - \frac{f_1}{f_2}) = 3 \frac{f_1}{f_2} -1

we substitute the values

              2n (1- [tex]\frac{5}{6.4}[/tex]) = 3 [tex]\frac{5}{6.4}[/tex] -1

              2n 0.21875 = 1.34375

              n = 1.34375 / 2 0.21875

              n = 3

remember that n must be an integer.

We use one of the equations to find the length of the Tunal

                  L = (2n + 1) \frac{v}{4f_1}

                  L = (2 3 + 1) [tex]\frac{333}{4 \ 5.0}[/tex]

                  L = 116.55 m

Answer:

0.1 second

Explanation:

We are given;

Power; P = 350 W

Force; F = 7 N

Distance; d = 5 m

Formula for power is;

P = workdone/time taken

Workdone = F × d

Thus;

350 = (7 × 5)/t

t = 35/350

t = 0.1 second

Answer:

Visible light waves

Explanation:

The answer is 18000 J

I hope this helps!^^ , if you need the work to be shown please tell me, I hope you have a great day!^^

Answer:

25 miles per hour

Explanation:

It was 20 miles per hour the next day.  We don't have enough information to calculate the average speed for the whole trip.

Answer:

E = 1.7 10² PJ

Explanation:

Let's use the special relativity relations, specifically the energy of a body is

           E = γ mc²

           γ = [tex]\sqrt{ 1 - \frac{v^2}{c^2} }[/tex]

where m is the rest mass

for that case they tell us that the speed of the body is 87% of the speed of light

                  = 0.87

let's calculate

         γ = [tex]\sqrt{1 - 0.87^2}[/tex]

         γ = 0.49305

let's reduce the mass of the jack to SI units

         W = 8.5 lb (4.448 N / 1lb) = 37.808 N

         W = mg

         m = W / g

         m = 37.808 / 9.8

         m = 3.86 kg

let's look for energy

          E = 3.86 (3 10⁸ )² 0.49305

          E = 1.7 10¹⁷ J

let's reduce to take PJ

          E = 1.7 10¹⁷ J ([tex]\frac{ 1 PJ}{10^{15} J}[/tex] )

          E = 1.7 10² PJ

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

Answer:

[tex]473.56\ \text{J}[/tex]

Explanation:

T = Time taken to complete one revolution = 0.3 s

m = Mass of sphere = 15 kg

r = Radius of sphere = 0.6 m

I = Moment of inertia of sphere = [tex]\dfrac{2}{5}mr^2[/tex]

Angular speed of the sphere is

[tex]\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.3}\\\Rightarrow \omega=20.94\ \text{rad/s}[/tex]

Rotational kinetic energy is given by

[tex]K_r=\dfrac{1}{2}I\omega^2=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2\\\Rightarrow K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 15\times 0.6^2\times 20.94^2\\\Rightarrow K_r=473.56\ \text{J}[/tex]

The rotational kinetic energy of the sphere is [tex]473.56\ \text{J}[/tex].

B

Explanation:

lipids contains fat

hope it helps

Answer:

A.Potential

B.Kinetic

C.Gravitational

Explanation:

I took the test

there you go, 15N. I hope this helps

Answer:

The answer is "[tex]\bold{18 \ \frac{km}{s}}[/tex]"

Explanation:

Its concern is not whether star speed is significantly lower than the light speed. Taking into consideration the relativistic tempo (small speed star)

[tex]\to \frac{\Delta \lambda}{\lambda} = \frac{v}{c}\\\\\to v = \frac{\Delta \lambda}{\lambda} \left (c \right ) \\\\[/tex]

       [tex]= \left ( \frac{0.04}{656.46} \right ) (3 \times 10^8)\\\\ = 18280 \ \frac{m}{s} \approx 18 \ \frac{km}{s}[/tex]

Answer:

a) ΔV ’= 1.66 10¹ V= 16.6 V,  b)  U = 55.64 10⁻¹² J,  c)    U_f = 1.18 10⁻¹⁰ J

d)     W = 6.236 10⁻¹¹ J

Explanation:

Capacitance can be found for a parallel plate capacitor

          C = ε₀  [tex]\frac{A}{d}[/tex]  

Let's reduce the magnitudes to the SI system

           A = 9.30 cm² (1 m / 10² cm) 2 = 9.30 10⁻⁴ m²

           c = 4.50 mm (1 m / 1000 mm) = 4.50 10⁻³ m

          Co = 8.85 10⁻¹²    9.30 10⁻⁴ /4.50 10⁻³

          Co = 1.829 10⁻¹² F

when the plates separate at d = 9.60 10⁻³ m, the capcitance changes to

          C = ε₀ \frac{A}{d_1}

          C = 8.85 10⁻¹² 9.30 10⁻⁴/9.60 10⁻³

          C = 8.57 10⁻¹³ F

a) the potential difference

            C =

since the capacitor is not discharged, let's look for the initial charge

            Co = \frac{Q}{ \Delta V}

             Q = C₀ ΔV

              Q = 1.829 10⁻¹² 7.80

             Q = 14.2662 10⁻¹² C

when the condensate plates are separated

             C = \frac{Q}{ \Delta V' }

              ΔV ’= Q / C

              ΔV ’= 14.266 10⁻¹² / 8.57 10⁻¹³

              ΔV ’= 1.66 10¹ V= 16.6 V

b) the stored energy is

             U = ½ C ΔV²

for initial separation

              U = ½ C₀ ΔV²

             U = ½ 1.829 10⁻¹² 7.80²

              U = 55.64 10⁻¹² J

c) The energy for end separation;

               U_f = ½ C DV’2

                U_f = ½ 8.57 10⁻¹³ 16,6²2

                U_f = 1.18 10⁻¹⁰ J

d) The work

as there are no losses, the work is equal to the variation of the energy

                W = ΔU = U_f -U₀

                 W = 1.18 10⁻¹⁰ - 55.64 10-12

                 W = 6.236 10⁻¹¹ J

Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

               sin φ = z / r

                Cos φ = r_p / r

               z = r sin φ

               r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

               r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

               r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

               r = √126

               r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

                φ = sin⁻¹ z / r

                φ = sin⁻¹ ( [tex]\frac{-7+4}{11.23}[/tex] )

                φ = 15.5º

Let's find the projection of the position vector (r_p)

                r_p = r cos φ

                r_p = 11.23 cos 15.5

                r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

                 cos θ = x / r_p

                 θ = cos⁻¹ x / r_p

                 θ = cos⁻¹ ( [tex]\frac{14-8}{10.82}[/tex])  

                 θ = 56.3º

taking the angles we can decompose the initial velocity.

               sin φ = v_z / v₀

               cos φ = v_p / v₀

               v_z = v₀ sin φ

               v_z = 5.5 sin 15.5 = 1.47 m / z

               v_p = vo cos φ

               v_p = 5.5 cos 15.5 = 5.30 m / s

               cos θ = vₓ / v_p

                sin θ = v_y / v_p

                vₓ = v_p cos θ

                v_y = v_p sin θ

                vₓ = 5.30 cos 56.3 = 2.94 m / s

                v_y = 5.30 sin 56.3 = 4.41 m / s

we already have the components of the initial velocity

                v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

                Fₓx = m aₓ₁

                aₓ₁ = Fₓ / m

                aₓ₁ = 220/100

                aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

                 vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

let's calculate

                 vₓ₁² = 2.94² + 2 2.20 (14-8)

                 vₓ₁ = √35.04

                 vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

                   aₓ₂ = Fₓ₂ / m

                   aₓ₂ = 100/100

                   aₓ₂ = 1 m / s²

final speed

                    v_{xf}²  = vₓ₁² + 2 aₓ₂ (x_f- x₁)

                    v_{xf}² = 5.92² + 2 1 (17-14)

                    v_{xf} =√41.05

                    v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

                      a_{y1} = F_y / m

                      a_{y1} = 460/100

                      a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

                      v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

                      v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

                      v_{y1} = √102.25

                       v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

                      a_{y2} = F_{y2} / m

                      a_{y2} = 260/100

                      a_{y2} = 2.60 m / s2

final speed

                     v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

                     v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

                      v_{yf} = √ 71.01

                      v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

first interval, relate (a_{z1})

                      a_{z1} = F_{z1} / m

                      a_{z1} = -200/100

                      a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

                    v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

                    v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

                    v_{z1} = √14.16

                    v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})

                    a_{z2} = F_{z2} / m

                    a_{z2} = 210/100

                    a_{z2} = 2.10 m / s2

final speed

                    v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)

                    v_{fz}² = 3.14² - 2 2.10 (-3 + 7)

                     v_{fz} = √6.94

                     v_{fz} = 2.63 m / s

speed is     v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Answer:

Option 4. is correct

Explanation:

A dashboard refers to a control panel that lies in front of the driver of the vehicle. Padded dashboards are designed to reduce injuries including face injuries and chest injuries to the driver as well as the front passenger on collisions.

Most modern vehicles have padded dashboards. This reduces collision injuries by increasing the time of impact.

Therefore,

Option 4. is correct

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         [tex]k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}[/tex]

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

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

we substitute

           q = [tex]\sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }[/tex]   m

            q = [tex]\sqrt{0.7419 \ 10^{-20}}[/tex]   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

this is a very small charge value so it should be easy to create in each one

Answer:

a)  A = 3.80 cm, b)   λ = 9.60 cm, c)  T = 1.75 10⁻² s, d)    v = 5.48 m / s

Explanation:

The wave is a way of transporting energy and moment without the need to transport the material. They are described by expressions of the type

           x = A sin (kx - wt)

where the amplitude A is the distance from the point of zero intensity to the maximum.

Frequency is the number of times the wave oscillates per unit of time

the wavelength is the distance necessary for the wave to start repeating.

a) In the exercise it tells us that the vertical distance from a machismo to a minimum that is worth 7.60 cm

when checking the definition of amplitude is from zero to a maximum, therefore the value given is twice the amplitude

          2A = 7.60

            A = 3.80 cm

b) the distance between a minimum and the next maximum is 4.80 cm

Using the definition of wavelength the given value corresponds to half wavelength

          λ/ 2 = 4.80

          λ = 9.60 cm

c) frequency and period are related

          f = 1 / T

          T = 1 / f

we calculate

           T = 1 / 57.1

           T = 0.0175 s

           T = 1.75 10⁻² s

d) the speed of the wave is related to the frequency and the wavelength

          v = λ f

          v = 0.0960 57.1

          v = 5.48 m / s