How many different uniquecurrents will there be in this circuit? How many different "branch" are there in this circuit where the currents could differ.

Answers

Answer 1

Answer:

hello your question is incomplete attached below is the missing circuit diagram

answer : 3 unique currents

Explanation:

From the circuit attached below it can be seen that there will be Three(3) unique currents flowing through this circuit and the Branches where this currents could differ is at the Three(3) resistors

How Many Different Uniquecurrents Will There Be In This Circuit? How Many Different "branch" Are There

Related Questions


Convert 200km/h to m/s​

Answers

Answer:

200 kilometers per hour =

55.556 meters per second

Explanation:

have a great day

Answer:

200km × 1000m/1km = 200000 m

1h × 60 mins/h × 60sec/min = 3600s

now you divide 200000m/3600s

= 55.55m/s

a decompression chamber used by deep sea divers has a volume of 10.1m^3 and operates at an internal pressure of 4.25 atm. What volume, in cubic meters would the air in the chamber occupy if it were at 1.00 atm pressure, assuming no tempreature change.

Answers

Answer:

V2=42.93cm3

Explanation:

According to Gay-Lussac's Law: "The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. With an increase in temperature, the pressure will go up."

P1V1=P2V2

Given data

P1=4.25atm

V1=10.1cm³

P2=1atm

V2=?

V2=P1V1/P2

V2=(4.25*10.1)/1

V2=42.93cm3

What happens if atoms lose energy during a change of state?

Answers

Answer: they become slower.

Explanation:  because they get pulled together.

Answer:

The atoms are pulled together by attractive forces and become more organized.

Explanation:

Atoms lose energy as a gas changes to a solid. ... Atoms gain energy as a solid changes to a liquid. If atoms energy during a change of state, they are pulled together by attractive forces and become more organized.

Reference by Quizlet.

s this statement true or false?

Hurricanes are classified by three stages in which updrafts of air billow up, updrafts and downdrafts swirl the wind up and down, then downdrafts cause clouds to come apart.


true

false

Answers

Answer:

True

Explanation:

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

When the momentum of an object increases with respect to time, what is true of the net force acting on it?

a. It is zero, because the net force is equal to the rate of change of the momentum.
b. It is zero, because the net force is equal to the product of the momentum and the time interval.
c. It is nonzero, because the net force is equal to the rate of change of the momentum.
d. It is nonzero, because the net force is equal to the product of the momentum and the time interval.

Answers

Answer:

C. It is nonzero, because the net force is equal to the rate of change of the momentum

Explanation:

Momentum of an object is given as;

P = mv = ft

f = mv/t

or f = P/t

where;

P is the momentum of the object

m is mass of the object

v is velocity of the object

f is the applied force on the object

t is time

From the given equation above; the net force acting on the object is equal to the rate of change of the momentum, thus it is non-zero.

Therefore, the correct option is "C"

C. It is nonzero, because the net force is equal to the rate of change of the momentum.

The net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.

The problem is based on the impulse-momentum concept. As per the impulse -momentum concept, the magnitude of impulse produced by an object is equal to the change in momentum. Then the expression is,

[tex]I = \Delta P[/tex]

here,

I is the magnitude of impulse. And its value is,

[tex]I = F \times t[/tex]

t is the impact time or time interval.

F is the magnitude of net force.

[tex]\Delta P[/tex] is the change in momentum.

Solving as,

[tex]I= \Delta P\\\\F \times t= \Delta P\\\\F = \dfrac{\Delta P}{t}[/tex]

Clearly, the net force is dependent of the change in momentum with respect to time. So, on increasing the momentum change with respect to time, the net force will also increase with non-zero value.

Thus, we can conclude that the net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.

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State two function of the mass M placed on the metre rule

Answers

correct answer is mass media is correct answer OK

What height does a frictionless playground slide need so that a 35 kg child reaches the bottom at a speed of 6.5 m/s?

Answers

Answer:

h = 2.15 m

Explanation:

Given that,

Mass of the child, m = 35 kg

The child reaches the bottom at a speed of 6.5 m/s

We need to find the height slide by a frictionless playground slide. Let it is equal to h. Using the conservation of energy to find it as follows :

[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(6.5)^2}{2\times 9.8}\\\\h=2.15\ m[/tex]

So, it will slide to a height of 2.15 m

The height slid by the frictionless playground can be calculated by equating the kinetic and potential energy equations which is 2.16 meters

Mass of child = 35 kg Velocity, v = 6.5 m/s g = 9.8 m/

K.E = P.E

K. E = 0.5mv² ; P.E = mgh

0.5mv² = mgh

gh = 0.5v²

h = [(0.5v²) ÷ g]

h = [(0.5 × 6.5²) ÷ 9.8]

h = (21.125 ÷ 9.8)

h = 2.155 m

Therefore, the height slid by a frictionless playground is 2.16 meters

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If the rock is kicked at initial velocity of 12 m/s from a height of 106 m above the ground what horizontal distance does the rock travel before striking the ground?

Answers

Answer:

74 m.

Explanation:

From the question given above, the following data were obtained:

Height (h) = 186 m

Initial velocity (u) = 12 m/s

Horizontal distance (d) =?

Next, we shall determine the time taken for the rock to get to the ground. This can be obtained as follow:

Height (h) = 186 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

H = ½gt²

186 = ½ × 9.8 × t²

186 = 4.9 × t²

Divide both side by 4.9

t² = 186/4.9

Take the square root of both side

t = √(186/4.9)

t = 6.16 s

Thus, the time taken for the rock to get to the ground is 6.16 s

Finally, we shall determine the horizontal distance travelled by the rock as follow:

Initial velocity (u) = 12 m/s

Time (t) = 6.16 s

Horizontal distance (d) =?

Horizontal distance (d) = Initial velocity (u) × Time (t)

d = u × t

d = 12 × 6.16

d = 73.92 ≈ 74 m

Therefore, the horizontal distance travelled by the rock is 74 m

1.A large beach ball weighs 4.0 N. One person pushes it with a force of 7.0 N due South while another person pushes it 5.0 N due East. Find the acceleration on the beach ball.

2. What is the weight of a 70 kg astronaut on the earth, on the moon, (g=1.6 m/s2), on Venus (g = 18.7 m/s2) and in outer space traveling at a constant velocity

Answers

Answer:

1. force applied southward = -4 j

force applied eastward = 5 i

total force applied = 5i - 4j

magnitude of total force applied = √(5)²+(-4)²

magnitude of total force applied = √25 + 16 = √41

magnitude of total force applied = 6.4N

But the beach ball also weighs 4 N,

which means that a force of 4N is required to overcome the inertia of the ball

Net force applied on the ball = total force applied - force applied by inertia

Net force applied on the ball = 6.4 - 4

Net force applied on the ball = 2.4 N

Mass of the ball:

Mass of the ball = weight of the ball / gravitational constant

Mass of the ball = 4 / 9.8 = 0.4 kg

Acceleration of Ball:

from newton's second law of motion:

F = ma

replacing the variables

2.4 = 0.4 * a          (where a is the acceleration of the ball)

a = 2.4/0.4

a = 6 m/s²

2. Mass of astronaut = 70 kg

Weight of Earth:

Weight = Mass * acceleration due to gravity

Weight = 70 * 9.8

Weight = 686 N

Weight on Moon:

Weight = Mass * acceleration due to gravity

Weight = 70 * 1.6                             (we are given that g = 1.6 on moon)

Weight = 112 N

Weight on Venus:

Weight = Mass * acceleration due to gravity

Weight = 70 * 18.7                         (we are given that g = 18.7 on Venus)

Weight = 1309 N

will give brainliest pls help

Answers

Answer:

a and c

Explanation:

119) You throw a rock horizontally off a cliff with a speed of 20 m/s and no significant air resistance. After 2.0 s, the magnitude of the velocity of the rock is closest to

Answers

Answer: 44.54 m/s

Explanation:

Let's break the problem into horizontal and vertical.

Horizontal:

Here we do not have any force, only a constant horizontal velocity, so we can write:

Vx(t) = 20m/s.

Vertical:

Here we have the gravitational acceleration acting on the rock, then we can write:

Ay(t) = -9.8m/s^2

For the vertical velocity, we need to integrate over time and get:

Vy(t) = (-9.8m/s^2)*t + V0

Where V0 is the initial vertical velocity, but the rock was thrown horizontally, so we do not have an initial velocity in the vertical axis.

Vy(t) = (-9.8m/s^2)*t.

Ok, now we want to know the magnitude of the velocity when t = 2.0s.

The magnitude will be equal to:

V = √( Vx(2s)^2 + Vy(2s)^2)

   = √(  (20m/s)^2 + (-2s*9.8m/s^2)^2) = 44.54 m/s

The magnitude of the velocity after 2 seconds is around 44.54 m/s

   

A small, 300 gg cart is moving at 1.30 m/sm/s on a frictionless track when it collides with a larger, 4.00 kgkg cart at rest. After the collision, the small cart recoils at 0.860 m/sm/s . Part A What is the speed of the large cart after the collision

Answers

Answer:

v₂f = 0.16 m/s (to the right)

Explanation:

Assuming no external forces acting during the collision, as stated by Newton's 2nd law, total momentum must be conserved.The initial momentum is due to the small cart only, because the big one is at rest.Assuming that the small cart is going to the right and taking this direction as the positive one, we can write the following expression for the initial momentum p₀ :

        [tex]p_{o} = m_{1} * v_{o1} = 0.3 kg * 1.3 m/s (1)[/tex]

The final momentum can be written as follows:

        [tex]p_{f} = m_{1} * v_{1f} + m_{2} * v_{2f} = 0.3 kg* (-0.86m/s) + 4.00 kg* v_{2f} (2)[/tex]

        As p₀ = pf, from (1) and (2), we can solve for v2f, as follows:

        [tex]v_{2f} = \frac{0.3kg* (1.3 + 0.86) m/s}{4.00kg} = \frac{0.3kg*2.16m/s}{4.00kg} = 0.16 m/s[/tex]

As the sign of v2f is positive, we conclude that it starts to move in the same direction that m1 was originally going (to the right), whilst m1 recoils, which means that after the collision, it bounces back from the larger mass.

If the ball is released from rest at a height of 0.63 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track

Answers

Answer:

When the ball is on the frictionless side of the track , the angular speed is 89.7 rad/s.

Explanation:

Consider the ball is a solid sphere of radius 3.8 cm and mass 0.14 kg .

Given ,  mass, m=0.14 kg

Ball is released from rest at a height of, h= 0.83 m

Solid sphere of radius, R = 3.8 cm

                                       =0.038 m

From the conservation of energy

                          ΔK  = ΔU  

                [tex]\frac{1}{2}mv^2 +\frac{1}{2} I\omega^2=mgh[/tex]

Here , [tex]I=\frac{2}{5} MR^2 , v= R \omega[/tex]

  [tex]\frac{1}{2}mv^2\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R^2} )=mgh[/tex]

  [tex]\frac{1}{2} [v^2+\frac{2}{5}v^2]= gh[/tex]

[tex]\frac{7}{10} v^2=gh[/tex]

[tex]0.7v^2=gh[/tex]

 v=[tex]\sqrt{[gh/(0.7)][/tex]

=[tex]\sqrt{ [(9.8 m/s^2)(0.83 m) / (0.7) ][/tex]

= 3.408 m/s

Hence, angular speed when it is on the frictionless side of the track,

[tex]\omega=\frac{v}{R}[/tex]

   = (3.408 m/s)/(0.038 m)

[tex]\omega[/tex]   =  89.7 rad/s

Hence , the angular speed is 89.7 rad/s

The microwave background radiation is observed at a wavelength of 1.9 millimeters. The red shift of these photons is observed to be raoughly 1100. What would be the emitted wavelength?

Answers

Answer:

1730nm

Explanation:

Observed wavelength λ= 1.9 millimeters

Red shift of photons = 1100

Emitted wavelength = λo

Red shift = λ - λo/ λo = 1100

= [(1.9x10^-3 - λo)/ λo] = 1100

= 1.9x10^-3 - λo = 1100 λo

= 1.9x10^-3 = 1100 λo + λo

= 1.9x10^-3 = 1101 λo

We divide through by 1101 to get the value of lambda

1.9x10^-3/1101 = λo

λo = 0.000001726

λo = 1726nm

This is approximately

λo = 1730nm

Thank you!

A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center. What is the ratio of the tangential acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod

Answers

Answer:

Explanation:

angular acceleration ω = 8 rad /s²

tangential acceleration a = angular acceleration x radius of circle of rotation

For a point on the end of the rod , radius of circle = L .

tangential acceleration = 8 L .

For  a point a distance L/2 from the end of the rod, radius of circle = L /2 .

tangential acceleration = 8 L/2 = 4 L  .

It is so because angular acceleration will be same for this point , only radius changes .

required ratio = 8L / 4L

= 2.

Jack drops a stone from rest off of the top of a bridge that is 23.4 m above the ground. After the stone falls 7.2 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock

Answers

Explanation:dropped stone hits water:

24.4-4.9t^2=0

t=2.23

stone has fallen 6.6m:

4.9t^2 = 6.6

t = 1.16

So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water:

24+1.07v-4.9*1.07^2 = 0

v = -17.19 m/s

Both the rocks hit the water at the exact same time, and the initial velocity of Jill's rock will be -19.28 m/s.

What is velocity?

A vector measurement of the rate and direction of motion is what is meant by the term "velocity." Simply said, velocity is the rate of movement in a single direction. Velocity may be used to gauge both the speed of a rocket launching into space and the speed of a car travelling north on a busy freeway.

According to the question, initial velocity will be 0.

By using equation of motion :

Let Jill is x and Jack is y.

s=ut+1/2at² where,

a is acceleration,

u is initial velocity and,

t is the time period.

23.4=(-1/2)(9.8)×t²

t(y)=[tex]\sqrt{(23.4)/(4.9)}[/tex]

=2.185 seconds.

Time taken by Jack's stone that it travel the first 7.2 meters.

t=[tex]\sqrt{(2*7.2)/9.8}[/tex] =1.21 seconds.

So, time taken by stone thrown by jack,

t(Jill)=t(jack)-t

=23.4=-ut(x)-1/2(gt²)(y)

-23.4=(-u×0.975)-(1/2×9.8×(0.975)²)

u= -(23.4-4.6/0.975)

u=-19.28 m/s

Hence, the initial velocity of the Jill's stone will be -19.28 m/s.

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a 214kg boat is sinking in the ocean.its acceleration towards the sea floor is 6.12 m/s^2. the force of gravity that draws the boat down is partially offset by the buoyant force of the water, so what is the force pulling on the boat as it sinks?

Answers

The weight of the boat is 214 x 9.8 = 2097 N
The net force downwards is 214 x 6.12 = 1310 N
The upwards force is therefore 2097 - 1310 = 787 N upwards

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 4.40 rev/s; 60.0 revolutions later, its angular speed is 15.0 rev/s. Calculate (a) the angular acceleration (rev/s2), (b) the time required to complete the 60.0 revolutions, (c) the time required to reach the 4.40 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 4.40 rev/s angular speed.

Answers

Answer:

The answer is below

Explanation:

a) Using the formula:

[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=15\ rev/s,w_o=initial\ velocity=4.4\ rev/s, \\\theta=distance=60\ rev\\\\Substituting:\\\\15^2=4.4^2+2(60)\alpha\\\\2(60)\alpha=15^2-4.4^2\\\\2(60)\alpha=205.64\\\\\alpha=1.71\ rev/s^2[/tex]

b) The disk is initially at rest. Using the formula:

[tex]\theta=\omega_ot+\frac{1}{2}\alpha t^2 \\\\but\ \omega_o=0(rest), \theta=60\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\60=0+\frac{1}{2}(1.71) t^2\\\\t=8.4\ s[/tex]

c)

[tex]\omega=\omega_o+\alpha t \\\\but\ \omega_o=0(rest), \omega=4.4 \ rev/s\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\4.4=1.71t\\\\t=2.6\ s[/tex]

d)

[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=4.4\ rev/s,w_o=initial\ velocity=0\ rev/s, \\\alpha=1.71\ rev/s^2\\\\Substituting:\\\\4.4^2=0+2(1.71)\theta\\\\19.36=3.42\theta\\\\\theta=5.66\ rev[/tex]

ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?

Answers

Answer:

A)90°

Explanation:

ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?

(a) What is the angle between and the coil's magnetic dipole moment.

The angle between and the coil's magnetic dipole moment is 90°, because it is perpendicular to the field since there is maximum torgue.

(b) What is the number of turns in the coil?

differentiate between speed and velocity​

Answers

Explanation:

Speed - The rate at which something moves

Velocity - The speed of something in a specific direction

Velocity is kind of a specific type of speed.

Speed,is a scalar quantity, and is the rate at which an object covers a distance. Speed is independant of direction.
velocity is a vector quantity. it is direction-dependant, and is the rate at which the position changes.

A Ferris wheel rotates at an angular velocity of 0.25 rad/s. Starting from rest, it reaches its operating speed with an average angular acceleration of 0.027 rad/s2. How long does it take the wheel to come up to operating speed?

Answers

Answer:

t = 9.25 s

Explanation:

Given that,

Initial angular velocity, [tex]\omega_o=0[/tex] (at rest)

Final angular velocity, [tex]\omega_o=0.25\ rad/s[/tex]

Angular acceleration, [tex]\alpha =0.027\ rad/s^2[/tex]

We need to find the time it take the wheel to come up to operating speed. We know that the angular acceleration in terms of angular speed is given by :

[tex]\alpha =\dfrac{\omega_f-\omega_o}{t}\\\\t=\dfrac{\omega_f-\omega_o}{\alpha }\\\\t=\dfrac{0.25-0}{0.027}\\\\t=9.25\ s[/tex]

So, it will reach up to the operating speed in 9.25 s.

A particle with charge q = +5.00 C initially moves at v = (1.00 î + 7.00 ĵ ) m/s. If it encounters a magnetic field B = 10 k, find the magnetic force vector on the particle.

Answers

Given :

Force on object moving with constant velocity due to magnetic field is given by :

[tex]\vec{F}=q\vec{v}\times \vec{B}\\\\\vec{F}=-5[ ( \hat{i} + 7\hat{j}) \times ( 10\hat{k})]\\\\\vec{F}=-5[ 10(-\hat{j})+70(\hat{i})]\\\\\vec{F}=50\hat{j}-350\hat{i}\ N[/tex]

Now,

[tex]F = \sqrt{50^2+350^2}\\\\F=353.55\ N[/tex]

Hence, this is the required solution.

A particle with charge q = +5.00 C that is moving with some velocity, then the magnetic force vector on the particle will be 353.55 N.

What is a vector?

The phrase "vector" in physics and mathematics is used popularly to refer to some values that cannot be stated by a simple integer (a scalar) or to certain elements of high - dimensional space.

For things like displacements, forces, and velocity can have both a magnitude and direction, vectors were initially introduced in geometry and physics. Similar to how lengths, masses, and time are represented by real numbers, these quantities were represented by geometric vectors.

In some contexts, tuples—finite sequences of numbers with a definite length—are sometimes referred to as vectors.

From the data given in the question,

F = qv × B

F = -5 [(î +7 ĵ) ×(10 k)]

F = 50 ĵ - 350 î N

Now, calculate the force,

F = √50² + 350²

F = 353.55 N.

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A tuna jumps out of the water with an initial velocity of 44 feet per second

Answers

Answer:

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Explanation:

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You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the
head and knocks you to the ground. From your handy dandy tourist guidebook, you find that the
height of the Eiffel Tower is 300.5 m. If you neglect air resistance, calculate how many seconds
the croissant dropped before it tagged you on the head.
What is the answer

Answers

s = 1/2 a t^2
300.5 = 1/2 x 9.8 x t^2
t = square root ( 61.3 )
t = 7.8 seconds

If we put a flammable material in a point where many light rays converge we can set it on fire. With what type of spherical mirror would it be possible to do this

Answers

Answer:

Concave mirror

Explanation:

A concave mirror is a type that has the property of converging rays of light that are parallel to the principal axis to a point of focus. Thus, for the convergence of many rays of light (i.e a beam of light) as required in the given question, a concave mirror would serve the purpose. Curved mirrors have various applications in optics.

Calculate the linear acceleration of a car, the 0.300-m radius tires of which have an angular acceleration of 17.0 rad/s2. Assume no slippage and give your answer in m/s2.

Answers

Answer:

5.1m/s^2

Explanation:

In this problem we want to solve for the linear acceleration

Given data

radius r= 0.3m

angular acceleration α=17rad/s^2

we know that the expression relating angular acceleration and linear acceleration is given as

a=rα

Substituting we have

a=0.3*17

a=5.1m/s^2

Two trains travel from New Orleans to Memphis in 5 h. The first train travels at a constant velocity of 80 mph, but the velocity of the second train varies. What was the second train's average velocity during the trip

Answers

Answer:

Explanation:

The first train travels at a constant velocity of 80 mph and it takes 5 h to cover the  distance so distance between two station = 80 x 5 = 400 m

second train too takes 5 h to cover the same distance so its average velocity

= total distance / total time

= 400 / 5 = 80 m /h

average velocity = 80 m /h

Which of The following statements best describes the objects motion between zero and four seconds?

Answers

Answer:

A OR C

Nearly A.

Explanation:

How much is the spring stretched, in meters, by an object with a mass of 0.49 kg that is hanging from the spring at rest

Answers

Answer:

1.6m

Explanation:

Using the spring equation, which is as follows:

F = Kx

Where; F = force applied on the spring (N)

K = spring constant (3kg/s²)

x = extension of spring (m)

However, Force applied by object = mass × acceleration (9.8m/s²)

F = 0.49kg × 9.8m/s²

F = 4.802N

Since, F = 4.802N

F = Kx

4.802 = 3 × x

4.802 = 3x

x = 4.802/3

x = 1.6006

x = 1.6m

The spring is stretched i.e. the extension of the spring, by 1.6m

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