A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the container the pressure is 116 kPa.
(a) What is the depth of the fluid?
(b) Find the pressure at the bottom of the container after an additional 2.05 X 10^-3 m^3 of this fluid is added to the container. Assume that no fluid spills out of the container.

Answers

Answer 1
The right answer is (b)

Related Questions

A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory

Answers

Answer:

Yes

Explanation:

speed of truck = 20 km/h

Initially the car at rest.

maximum velocity of car = 40 km/h

When the truck and the car collide, the momentum of the truck transferred to car.

So, the car can attain the speed of 40 km/h.

A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find

(i) Final velocity

(ii)The time taken​

Answers

Answer:

(I)

[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]

(ii)

[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g

Answers

Answer:

w = 1,066 rad / s

Explanation:

For this exercise we use Newton's second law

         F = m a

the centripetal acceleration is

         a = w² r

indicate that the force is the mass of the body times the acceleration

        F = m 0.58g = m 0.58 9.8

        F = 5.684 m

we substitute

       5.684 m = m w² r

       w = [tex]\sqrt{5.684/r}[/tex]

To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m

       w = [tex]\sqrt{ 5.684/5}[/tex]

       w = 1,066 rad / s

An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?

Answers

Answer:

0.5 V

Explanation:

The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.

Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.

Answers

Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).

Explanation:

There are different modes of heat energy transfer which includes:

--> conduction

--> Radiation and

--> Convection

CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.

Convection in gases is very common and gas expands more than liquid when subjected to high temperature.

--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).

-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.

-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.

--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.

But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.

Drag the titles to the correct boxes to complete the pairs.

Answers

Can you input a picture??

Wood is an example of
A. Metalloid
B. Insulator
C. Nonmetal
D. Conductor

Answers

The Answer is b: insulator

Electrical charges are of two types. True False

Answers

Answer:

Electrical charges r of 2 types its true.they are positive and negative.

hope it helps.stay safe healthy and happy..

Answer: Think its true

A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster?

Answers

Answer:

When an object is dropped, the "principal" force that acts on that object is the gravitational force.

Thus, in the absence of air resistance and such, the acceleration of the object will be equal to the gravitational acceleration:

g = 9.8m/s^2

So, when we drop objects in the moon (where there is no air) the acceleration of every object will be exactly the same. (so there is no dependence in the mass or shape of the object)

Thus, if we drop a coin and a feather in the moon, both objects will fall with the same acceleration, and then both objects will hit the ground at the same time.

But if we are in Earth, we can not ignore the air resistance (a force that acts in the opposite direction than the movement of the object)

And this force depends on the shape and mass of the object (for example, something with a really larger surface and really thin, like a sheet of paper will be more affected by this force than a small rock)

Then here, when the air resistance applies, we should expect that the heavier and smaller object (the coin) to be less affected by this force, then the resistance that the coin experiences is smaller, then the coin falls "faster" than the feather.

Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm. ​

Answers

Answer:

Answer

Correct option is

A

5×10

−6

tesla

I=5A

x=0.2m

Magnetic field at a distance 0.2 m away from the wire.

B=

2πx

μ

0

I

=

2π×0.2

4π×10

−7

×5

=10×5×10

−7

=5×10

−6

tesla

Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:

a. 1,5 N

b. 15,0 N

c. 3,1 N

d. 5,2 N

Answers

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

Frequency, f = 120 Hz

wavelength = 0.6 m

Speed = frequency x wavelength

speed = 120 x 0.6 = 72 m/s

Let the tension is T.

Use the formula

[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]

Option (A) is correct.

A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing

Answers

Answer:

3.464 seconds.

Explanation:

We know that we can write the period (the time for a complete swing) of a pendulum as:

[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]

Where:

[tex]\pi = 3.14[/tex]

L is the length of the pendulum

g is the gravitational acceleration:

g = 9.8m/s^2

We know that the original period is of 2.00 s, then:

T = 2.00s

We can solve that for L, the original length:

[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]

So if we triple the length of the pendulum, we will have:

L' = 3*0.994m = 2.982m

The new period will be:

[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]

The new period will be 3.464 seconds.

Newspapers often talk about an energy crisis-about running out of certain energy sources in the not-so-distant future. About which kind of energy sources are they talking

Answers

Answer:

Nonrenewable energy

Explanation:

Renewable energy is also known as clean energy and it can be defined as a type of energy that are generated through natural sources or technology-based processes that are replenished constantly. Some examples of these natural sources are water (hydropower), wind (wind energy), sun (solar power), geothermal, biomass, waves etc.

Basically, a renewable energy source is sustainable and as such can not be exhausted.

On the other hand, a non-renewable energy refers to an energy source such as fossil fuels that takes a very long time to be created or their creation happened long ago and isn't likely to happen again e.g uranium.

For example, fossil fuels such as coal, oil, and natural gas, come from deep inside the Earth where they formed over millions of years ago.

In this scenario, the kind of energy the newspaper sources are talking about is a nonrenewable energy source because they are capable of being exhausted in the not-so-distant future.

b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide

Answers

Answer:

The two skaters move with a speed of 1.73 m/s after the collision in the right direction.

Explanation:

Given that,

The mas of skater 1, m₁ = 80 kg

The speed of skater 1, u₁ = 5 m/s (right)

The mass of skater 2, m₂ = 70 kg

The speed of skater 2, u₂ = -2 m/s (left)

Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]

Put all the values,

[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]

So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.

A 2090-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) =At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50m/s 2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?

Answers

Answer:

a) A = 1.50 m / s²,  B = 1.33 m/s³,  b) a = 12.1667 m / s²,

c)  I = M (1.5 t + 1.333 t²) ,  d)  ΔI = M 2.833   N

Explanation:

In this exercise give the expression for the speed of the rocket

         v (t) = A t + B t²

and the initial conditions

         a = 1.50 m / s² for t = 0 s

         v = 2.00 m / s for t = 1.00 s

a) it is asked to determine the constants.

Let's look for acceleration with its definition

         a = [tex]\frac{dv}{dt}[/tex]

         a = A + 2B t

we apply the first condition t = 0 s

         a = A

         A = 1.50 m / s²

we apply the second condition t = 1.00 s

          v = 1.5 1 + B 1²

          2 = 1.5 + B

          B = 2 / 1.5

          B = 1.33 m/s³

the equation remains

           v = 1.50 t + 1.333 t²

b) the acceleration for t = 4.00 s

           a = 1.50 + 1.333 2t

           a = 1.50 + 2.666 4

           a = 12.1667 m / s²

c) The thrust

           I = ∫ F dt = p_f - p₀

           

Newton's second law

          F = M a

          F = M (1.5 + 2 1.333 t) dt

           

we replace and integrate

         I = M ∫ (1.5 + 2.666 t) dt

         I = 1.5 t + 2.666 t²/2

         I = M (1.5 t + 1.333 t²) + cte

in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant

         cte = 0

         I = M (1.5 t + 1.333 t²)

d) the initial push

For this we must assume some small time interval, for example between

t = 0 s and t = 1 s

        ΔI = I_f - I₀

        ΔI = M (1.5 1 + 1.333 1²)

        ΔI = M 2.833   N

A satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit

Answers

Answer:

 T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Explanation:

Let's use Newton's second law

          F = ma

force is the universal force of attraction and acceleration is centripetal

          G m M / r² = m v² / r

          G M / r = v²

as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion

          v = d / T

the length of a circle is

          d = 2π r

we substitute

        G M / r = 4π² r² / T²

        T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]

the distance r is measured from the center of the Earth (Re), therefore

        r = Re + h

where h is the height from the planet's surface

let's calculate

         T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex]   (Re + h) ³

         T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]

         T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 412 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 412 K at constant pressure by doing work W2 while transferring energy Q2 by heat.

Required:
Find ÎEint, 1, Q1, and W1 for the process at constant volume.

Answers

Answer:

ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J

Explanation:

Given the data in the question;

T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol

since helium is monoatomic;

Cv = (3/2)R, Cp = (5/2)R

W₁ = 0 J [ at constant volume or ΔV = 0]

Now for the first cylinder; from the first law of thermodynamics;

Q₁ = ΔE[tex]_{int[/tex],₁ +  W₁

Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT

we substitute  

Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )

Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112

Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J

Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J

The atoms in your body are mostly empty space . And so are the atoms in any wall. Why then is your body unable to pass through walls ?

Answers

First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.

This was something from logic.

The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.

A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm.What is the spring constant k?How long is the spring when a 4.0 kg mass is suspended from it?

Answers

As the spring is stretched, it exerts an upward restoring force f. At maximum extension, Newton's second law gives

F = f - mg = 0   ==>   f = (2.0 kg) (9.8 m/s²) = 19.6 N

By Hooke's law, if k is the spring constant, then

f = kx   ==>   k = f/x = (19.6 N) / (0.15 m) ≈ 130 N/m

A 4.0 kg mass would cause the spring to exert a force of

f = (4.0 kg) (9.8 m/s²) = 39.2 N

which would result in the spring stretching a distance x such that

39.2 N = (130 N/m) x   ==>   x = (39.2 N) / (130 N/m) ≈ 0.30 m ≈ 30 cm

Please show steps as to how to solve this problem
Thank you!

Answers

Explanation:

Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:

[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]

[tex] -m_1gx + m_2gd_2 = 0[/tex]

[tex]m_1x = m_2d_2[/tex]

Solving for x,

[tex]x = \dfrac{m_2}{m_1}d_2[/tex]

[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]

A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as the velocity vector.

a. True
b. False

Answers

I believe it is False, only because the plane is Frictionless. Hope this helps, good luck.

Answer:

False

Explanation:

You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.

Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.

Answers

Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Explanation:

Let us assume that

[tex]q_{1} = q_{2} = +q[/tex]

[tex]q_{3} = -q[/tex]

As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).

Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r

Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]

where,

k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]

Substitute the values into above formula as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)

Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex]   ... (2)

As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.

Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb

Answers

Answer:

i₀ = V / R_i

Explanation:

For this exercise we use Ohm's law

         V = i R

          i = V / R

the equivalent resistance for

         [tex]\frac{1}{R_{eq}}[/tex] =  ∑ [tex]\frac{1}{R_i}[/tex]

if all the bulbs have the same resistance, there are N bulbs

         [tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]

         R_{eq} = R_i / N

we substitute

         i = N V / Ri

where i is the total current that passes through the parallel, the current in a branch is

         i₀ = i / N

         i₀ = V / R_i

0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b. 300 K c. 250 K d. 125 K​

Answers

Answer:

c. 250k

Explanation:

The temperature of the sink is approximately 250 K.

To find the temperature of the sink, we can use the formula for the efficiency of a heat engine:

Efficiency = 1 - (Temperature of Sink / Temperature of Source)

Given that the temperature of the source (T_source) is 500 K and the source energy (Q_source) is 2003 J, and the sink energy (Q_sink) is 100 J, we can rearrange the formula to solve for the temperature of the sink (T_sink):

Efficiency = (Q_source - Q_sink) / Q_source

Efficiency = (2003 J - 100 J) / 2003 J

Efficiency = 1903 J / 2003 J

Efficiency = 0.9497

Now, plug the efficiency back into the first equation to solve for T_sink:

0.9497 = 1 - (T_sink / 500 K)

T_sink / 500 K = 1 - 0.9497

T_sink / 500 K = 0.0503

Now, isolate T_sink:

T_sink = 0.0503 * 500 K

T_sink = 25.15 K

Since the temperature should be in Kelvin, we round down to the nearest whole number, which is 25 K. Thus, the temperature of the sink is approximately 250 K.

To learn more about sink energy, here

https://brainly.com/question/10483137

#SPJ2

A closely wound, circular coil with radius 2.70 cm has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is 0.0750 T

Answers

Answer:

Approximately 4.029 A

Explanation:

We can use the formula that the B field of a few loops all with current in same direction is permeability of free space (mu)* current * Number or loops divided by 2*radius. You are given B field, radius(convert into meters), number of loops and mu is 4pi * 10^-7. Solve for current and you get 4.029 Amperes.

a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?​

Answers

Answer:

Look at explanation

Explanation:

a)Only force acting on the object is gravity, so a=-g (consider up to be positive)

use: v^2=v0^2+2a(y-y0)

plug in givens, at max height v=0

0=400-19.6(H)

Solve for H

H= 20.41m

b) Use: y=y0+v0t+1/2at^2

Plug in givens

0=0+20t-4.9t^2

solve for t

t=4.08 seconds

c) v=v0+at

v=20-39.984= -19.984m/s

A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.

Answers

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, [tex]$\omega = \frac{1}{8}$[/tex]  rev/sec

                             [tex]$=\frac{2 \pi \times 7.5}{8}$[/tex]  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

[tex]$F=m.\omega^2.r$[/tex]

   [tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of
ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.

magnitude m/s
direction ° counterclockwise from the +x-axis

Answers

Answer:

a)    v = 517.99 m / s,  b) θ = 296.3º

Explanation:

This is an exercise in kinematics, we are going to solve each axis independently

X axis

the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ=  3670 m / s, let's use the relation

           vₓ = v₀ₓ + aₓ t

           v₀ₓ = vₓ - aₓ t

           v₀ₓ = 3670 - 5.10 670

           v₀ₓ = 253 m / s

Y axis  

the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after

t = 670 s

          v_y = v_{oy} + a_y t

          v_{oy} = v_y - a_y t

          v_oy} = 4378 - 7.30 670

          v_{oy}  = -513 m / s

to find the velocity modulus we use the Pythagorean theorem

          v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]

          v = [tex]\sqrt{253^2 +513^2}[/tex]

          v = 517.99 m / s

to find the direction we use trigonometry

         tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]

         θ'= tan⁻¹  [tex]\frac{voy}{voy}[/tex]  

         θ'= tan⁻¹ (-513/253)

         tea '= -63.7

the negative sign indicates that it is below the ax axis, in the fourth quadrant

to give this angle from the positive side of the axis ax

          θ = 360 -   θ  

          θ = 360 - 63.7

          θ = 296.3º

Which one of the following statements concerning resistors in "parallel" is true? Question 7 options: The voltage across each resistor is the same. The current through each resistor is the same. The total current through the resistors is the sum of the current through each resistor. The power dissipated by each resistor is the same.

Answers

Answer: The correct statement is:

--> The voltage across each resistor is the same.

Explanation:

RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:

--> Resistors in parallel and

--> Resistors in series

RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.

why is the water drawn from the bottom of the dam rather than the top?​

Answers

Answer:

because minerals can be gotten from the bottom

Explanation:

it's self explanatory

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