A 20 g "bouncy ball" is dropped from a height of 1.8 m. It rebounds from the ground with 80% of the speed it had just before it hit the ground. Assume that during the bounce the ground causes a constant force on the ball for 75 ms. What is the force applied to the ball by the ground in N?
The following are not correct: 0.513 N, 0.317 N, 0.121 N. Please show your work so I can understand!

' A ' looks like the best choice.

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation:

Answer:

A sound level of 96 dB is 4 times as intense as one of 90 dB

Explanation:

The formula of the intensity level of sound in decibels is given as follows:

Intensity Level = 10 log₁₀(I/I₀)

where,

I = Intensity of Sound

I₀ = Reference Intensity Level = 10⁻¹² W/m²

Therefore, for 96 dB sound level:

96 = 10 log₁₀(I₁/10⁻¹²)

log₁₀(I₁/10⁻¹²) = 96/10

I₁/10⁻¹² = 10^9.6

I₁ = (10⁻¹²)(4 x 10⁹)

I₁ = 0.004 W/m²

For 90 dB sound level:

90 = 10 log₁₀(I₂/10⁻¹²)

log₁₀(I₂/10⁻¹²) = 90/10

I₂/10⁻¹² = 10^9

I₂ = (10⁻¹²)(10⁹)

I₂ = 0.001 W/m²

Therefore,

I₁/I₂ = 0.004/0.001

I₁ = 4 I₂

Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.

With smaller gravitational forces and therefor less vertical acceleration, the projectile launched on the moon ... with the same initial speed and direction ...

-- climbs faster,

-- spends more time climbing,

-- reaches a higher peak,

-- falls slower,

-- spends more time falling, and

-- covers more horizontal distance

than the projectile launched on the Earth.

This is not because of air resistance.  It would be true even if there were no air resistance on the Earth.  It's entirely a gravity thing.  

Answer:

Light is less energetic than X-rays.

Explanation:

The electromagnetic spectrum refers to the range of wavelengths or frequencies over which electromagnetic radiation extends. It is the entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light. In the electromagnetic spectrum, the entire distribution of electromagnetic radiation is done according to their frequency or wavelength.

The energy of an electromagnetic wave depends on its frequency and wavelength. The shorter the wavelength, the greater the energy of the electromagnetic wave but the larger frequency, the greater the energy of the electromagnetic wave.

X-rays has a frequency of about 1×10^20 Hz compared to visible light of frequency of about 1×10^15 Hz. Hence X-rays, having a larger frequency, is more energetic than visible light.

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

[tex]U=(M+m)gh[/tex]       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]

The potential energy is:

[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

[tex]Mv_1+mv_2=(M+m)v[/tex]    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

[tex]M(0)+mv_2=(M+m)v[/tex]    

[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]    

The force on the cord after the impact is 2.59N

Answer:

Explanation:

Apply the law of conversion of energy

[tex]V_f = \sqrt{2gh}[/tex]

where,

h = height of which the bullet and block rise after impact

[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]

Therefore,

[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]

From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]

[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]

C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system

[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]

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Answer:

[tex]\tau = 1\ ms[/tex]

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

[tex]C = \epsilon_0 * area / distance[/tex]

Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]

So the capacitance is:

[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]

[tex]C = 10^{-12}\ F = 1\ pF[/tex]

The time constant of a rc-circuit is given by:

[tex]\tau = RC[/tex]

So we have that:

[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]

Answer:

The time taken is  [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Explanation:

From the question we are told that

     The speed of the current is  [tex]v__{R}}[/tex]

     The speed of the swimmer in direction of current is [tex]v__{S}}[/tex]

      The distance traveled by the swimmer is  [tex]D[/tex]

       The time taken to travel this distance is  [tex]t_{out}[/tex]

      The speed of the swimmer against  direction of current is  [tex]v__{s}}[/tex]

The resultant speed for downstream current is

       [tex]V_{r} = v__{S}} +v__{R}}[/tex]

The time taken can be mathematically represented as

      [tex]t_{out} = \frac{D}{V_{r}}[/tex]

      [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Answer:(a)

Explanation:

Student must have known that insulators can only be charged when they are rubbed against each other. In this process, one becomes electrically negative while other becomes electrically positive such that both have the same magnitude. The one which gains electrons becomes electrically negative due to the transfer of electrons while others lose the electron becomes positive due to the transfer of an electron to another body.

Answer:

Explanation:

From Newton's 2nd Law,

F = m×a

Where F is Force

m is mass

a is acceleration

Hence a= F/m

a= 16/8= 2m/s2

Complete Question

The complete question is shown on the first uploaded image

Answer:

The distance is [tex]r_2 = 4 \ AU[/tex]

Explanation:

From the question we are told that

    The size of Jupiter is  [tex]s_2 = 143,000 \ km[/tex]

    The  length of the International Space Station is [tex]r_1 = 380\ km[/tex]

    The  size of the International Space Station is  [tex]s_1 = 90 \ m =0.09 \ km[/tex]

The angular size where the same one night and this angular size is mathematically represented as

      [tex]\theta = \frac{s}{r}[/tex]

Since  [tex]\theta[/tex] is constant

        [tex]\frac{s_1}{r_1} = \frac{s_2}{r_2}[/tex]

substituting values

      [tex]\frac{0.09}{380} = \frac{143000}{r_2}[/tex]

=>   [tex]r_2 = 6.04 * 10^{9} \ km[/tex]

Now we are told to convert to AU and  1 AU  [tex]= 1.5 * 10^8 \ km[/tex]

  So

      [tex]r_2 = \frac{6.04 * 10^8}{1.5*10^{8}}[/tex]

      [tex]r_2 = 4 \ AU[/tex]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.

The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.

Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.

The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)

The horizontal distance traveled by particle is

x = ucosθt

t = x/ucosθ

The force in an electric field is F = qE...................(1)

where, q is charge , E is the strength of electric field

From, newton 2nd law of motion, Force F = ma.................(2)

Equating both the equations, we get

ma = qE

a = qE/m..................(3)

The vertical distance, y =usinθt - 1/2at²

From equation 3, we have

y = usinθt  -  1/2 (qE/m) t²

if y = 0, t = 2musinθ/(qE) = x / (ucosθ)

The electric field is represented as

Also, E = 2mu²×sinθ×cosθ/(xq)

Plug the values, we get

E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)

E = -4556.18 N/m

Thus, the electric field of the particle is  -4556.18 N/m.

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Answer:

30 minutes

Explanation:

Energy per time is constant, so:

E₁ / t₁ = E₂ / t₂

m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂

(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t

(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t

3/min = 90 / t

t = 30 min

Answer:

a. FALSE

b. FALSE

c. TRUTH

d. FALSE

e. FALSE

Explanation:

To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:

[tex]V=\frac{Q}{4\pi \epsilon_o R}[/tex]      inside the sphere

[tex]V'=\frac{Q}{4\pi \epsilon_o r}[/tex]      for r > R (outside the sphere)

R: radius of the sphere

ε0: dielectric permittivity of vacuum

Q: charge of the sphere

As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.

Hence, you can conclude:

a. The potential at the center of the sphere is zero. FALSE

b.The potential is lowest, but not zero, at the center of the sphere. FALSE

c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH

d. The potential at the center is the same as the potential at infinity. FALSE

e. The potential at the surface is higher than the potential at the center. FALSE

Answer:

A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)

B. T = 14859.5 N = 14.89 KN

Explanation:

A.

First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:

s₁ = Vi t + (0.5)at²

where,

s₁ = distance covered during accelerated downward motion = ?

Vi = initial speed = 0 m/s (since elevator is initially at rest)

t = time taken = 6 s

a = acceleration = 1.5 m/s²

Therefore,

s₁ = (0 m/s)(6 s) + (0.5)(1.5 m/s²)(6 s)²

s₁ = 4.5 m

also we find the final velocity using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (1.5 m/s²)(6 s)

Vf = 9 m/s

Now, the second part of downward motion is with constant velocity. So:

s₂ = vt

where,

s₂ = distance covered during constant speed downward motion = ?

v = Vf = 9 m/s

t = 6 s

Therefore,

s₂ = (9 m/s)(6 s)

s₂ = 54 m

Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:

s₃ = s₁ = 4.5 m

Therefore, the total distance covered by elevator is given by following equation:

s = s₁ + s₂ - s₃      (Downward motion taken positive)

s = 4.5 m + 54 m - 4.5 m

s = 54 m

Therefore, net motion of the elevator was 54 m downwards.

So the final floor will be:

Final Floor = Initial Floor - Distance Covered/Length of a floor

Final Floor = 29 - 54 m/4m

Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)

B.

The maximum tension will occur during the upward accelerated motion. It is given by the formula:

T = m(g + a)

where,

T = Max. Tension in Cable = ?

m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg

g = 9.8 m/s²

a = acceleration = 1.5 m/s²

Therefore,

T = (1315 kg)(9.8 m/s² + 1.5 m/s²)

T = 14859.5 N = 14.89 KN

Answer:

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Explanation:

The Free Body Diagram associated with the experiment is presented as attachment included below.

Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.

Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:

[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]

[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]

Where:

[tex]W[/tex] - Weight of the eraser, measured in newtons.

[tex]f[/tex] - Friction force, measured in newtons.

[tex]N[/tex] - Normal force, measured in newtons.

[tex]\theta[/tex] - Angle of the incline, measured in degrees.

The maximum allowable static friction force is:

[tex]f = \mu_{s} \cdot N[/tex]

Where:

[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

Likewise, the kinetic friction force is described by the following model:

[tex]f = \mu_{k} \cdot N[/tex]

Where:

[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])

In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:

[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]

[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]

But [tex]N = m\cdot g \cos \theta[/tex], so that:

[tex]\mu = \tan \theta[/tex]

Now, coefficients of static and kinetic friction are, respectively:

[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]

[tex]\mu_{s} \approx 0.716[/tex]

[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]

[tex]\mu_{k} \approx 0.596[/tex]

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Answer:

1) n = 4.47*10^12 photons

2) K = 0.25 eV

Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex]     (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

[tex]E_p=h\frac{c}{\lambda}[/tex]         (2)

c: speed of light = 3*10^8 m/s

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 600*10^-9 m

You replace the values of the parameters in the equation (2):

[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]

Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:

[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]

The number of photons is 4.47*10^12

2) The kinetic energy of the electrons emitted by the metal is given by the following formula:

[tex]K=E_p-\Phi[/tex]     (3)

Ep: energy of the photons

Φ: work function of the metal = 1.7 eV

You first convert the energy of the photons to eV:

[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]

You replace in the equation (3):

[tex]K=1.95eV-1.7eV=0.25eV[/tex]

The kinetic energy of the electrons emitted by the metal is 0.25 eV

(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]

(2). The maximum kinetic energy of the electron is 0.37eV.

(1). The power of light is given as,

            [tex]P=1.4*10^{-6}W[/tex]

    Energy is given as,

             [tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]

Number of photons per second is,

                    [tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]

(2). the maximum kinetic energy of the electron is,

              [tex]K.E=E-\phi[/tex]

Where [tex]\phi[/tex] is work function.

         [tex]K.E=2.07-1.7=0.37eV[/tex]

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Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

Answer:

  8.3 feet

Explanation:

The kinetic energy of the system on the ground is ...

  KE = Σ(1/2)(mv^2) = (1/2)(155)(24^2) +(1/2)(13)(0^2) = 44640 lb·ft²/s²

The potential energy at the highest point is the same:

  PE = mgh

  44640 = (155 +13)(32)h

  h = 44640/5376 = 8.30 . . . . feet

_____

We haven't worried too much about the conversion between pounds mass and pounds force. Whatever factor may be involved will divide out when computing the maximum height. We have used g=32 ft/s².

__

To achieve a 10 ft height, the running speed would need to be 26.34 ft/s, about 10% higher.

Answer:

2Ω

Explanation:

Ohm's law:

V = IR

10 V = (5 A) R

R = 2 Ω

Answer:

Energy is stored by a 50 ampere-minute 12

volt battery is approximately = 36,000 J = 36 kJ

Explanation:

Power in electrical circuits is given as

Power = IV

But power generally is defined as energy expended per unit time

Power = (Energy/time)

Energy = Power × Time

Energy = IV × Time

Energy = (I.t × V)

I.t = 50 Ampere-minute = 50 × 60 = 3000 Ampere-seconds

V = 12 V

Energy = 3,000 × 12 = 36,000 J = 36 kJ

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Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

[tex]distancia=\frac{1 segundo*90 km}{3600 segundos}[/tex]

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

[tex]distancia=\frac{0.025 km*1,000 metros}{1 km}[/tex]

distancia= 25 metros

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

La cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

Para obtener la velocidad de la motocicleta en un segundo, necesitaremos convertir 90 km / h en metros por segundo

Usando la tasa de conversión;

1000 m = 1 km

1 hora = 3600 segundos

[tex]\frac{90km}{hr} = \frac{90km \times 1000 m}{1km \times 3600s} \\\\\frac{90km}{hr} = \frac{90,000 m}{3600s} =25m/s \\[/tex]

Esto muestra que la cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

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Answer:

Salt

Explanation:

Salt plays a crucial role in maintaining human health. It is the main source of sodium and chloride ions in the human diet. Sodium is essential for the nerve and muscle function and is involved in the regulation of fluids in the body. Sodium also plays a role in the body's control of blood pressure and volume. Salt is harvested by seawater or brine is fed into large ponds of water and is drawn out through natural evaporation which allows the salt to be subsequently harvested.

Have a good day and stay safe!

Answer:

A) Fb = 671.3 N

B) The diver will sink.

Explanation:

A)

The buoyant force applied on an object by a fluid is given by the following formula:

Fb = Vρg

where,

Fb = Buoyant Force = ?

V = Volume of the water displaced by the object = 68.5 L = 0.0685 m³

ρ = Density of Water = 1000 kg/m³

g = 9.8 m/s²

Therefore,

Fb = (0.0685 m³)(1000 kg/m³)(9.8 m/s²)

Fb = 671.3 N

B)

Now, in order to find out whether the diver sinks or float, we need to find weight of the diver with gear.

W = mg = (71.8 kg)(9.8 m/s²)

W = 703.64 N

Since, W > Fb. Therefore, the downward force of weight will make the diver sink.

The diver will sink.

Answer:

WEIGHT ON MOON IS 0.2004N

Explanation:

mass of the body=120g=[tex]\frac{120}{1000}[/tex]kg=0.12kg (we will convert g into kg)

gravity on moon=1.67m/s²( to find the mass of anybody on another we should know its gravity)

as we know that (from the formula of weight)

weight=mass×gravity

w=mg

w=0.12kg²×1.67m/s²

w=0.2004N

Answer:

Potential difference is measured in volts

Explanation:

The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb.

Answer:

Your answer is A.) volts

Explanation:

Answer:

The correct option is;

- 4x

Explanation:

From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source

The inverse square law can be presented as follows;

[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]

As the distance, r, increases, the surface it covers also increases by the power of 2

Therefore, where the distance increases from r to 2·r, we have;

When, I, remain constant

[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]

The surface increases to 4·S by the inverse square law

Therefore, the correct option is 4 × x.

Answer: B, Companies passed on production and transportation costs to consumers

Explanation:

A higher oil price occurred when companies passed on production and transportation costs to consumers.

The oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.

The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.

Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.

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