3) Given R, = 4 ohms and R, = 5 ohm and V = 18 volts
d) Find the total resistance: I
Find the current in the circuit:
f) Find the voltage drop in each resistor:

Answers

Answer 1

Answer:

d) the total resistance is 4 + 5 = 9 ( ohm)

e) the current in the circuit is 18/9 = 2(A)

f) the voltage in R = 4 ohm is 2.4 = 8 (V)

the voltage in R = 5 ohm is 18 - 8 = 10 (V)


Related Questions

A 2.0 kg block of ice with a speed of 8.0 m/s makes an elastic collision with another block of ice that is at rest. The first block of ice proceeds in the same direction as it did initially, but with a speed of 2.0 m/s. What is the mass of the second block? (Hint: Use the conservation of kinetic energy to solve for the second unknown variable.)

Answers

Answer:

6kg

Explanation:

According to conservation of kinetic energy

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses of the bodies

u1 and u2 are the initial velocities

v is the final velocity

Given

m1 =2kg

u1 = 8.0m/s

m2 = ?

u2 = 0m/s (second ice at rest)

v = 2.0m/s

Substitute into the formula

2(8)+m2(0) = (2+m2)(2)

16+0 = 4+2m2

16-4= 2m2

12 = 2m2

m2 =12/2

m2 = 6kg

Hence the mass of the second block is 6kg

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0o. (b) What is the minimum coefficient of friction needed for a frightened dri

Answers

Answer:

a) The ideal speed = 16.21 m/s

b) Minimum co-efficient of friction = 0.216

Explanation:

From the given information:

The ideal speed can be determined by considering the centrifugal force component and the gravity component.

[tex]\dfrac{mv^2}{r}cos \theta = mg sin \theta[/tex]

[tex]v = \sqrt {gr \ tan \theta}[/tex]

[tex]= \sqrt{(9.8 \ m/s^2) (100) \ tan 15^0}[/tex]

= 16.21 m/s

(b)

Let assume that it requires 25 km/h to take the same curve.

Then, using the equilibrium conditions;

[tex]mg \ sin \theta = \dfrac{mv^2}{r} cos \theta + \mu ((\dfrac{mv^2}{r}) sin \theta + mg cos \theta)[/tex]

[tex]\mu = \dfrac{mg sin \theta - \dfrac{mv^2}{r} cos \theta }{((\dfrac{mv^2}{r}) sin \theta + mg cos \theta) }[/tex]

[tex]\mu = \dfrac{g sin \theta - \dfrac{ v^2}{r} cos \theta }{((\dfrac{v^2}{r}) sin \theta + g cos \theta) }[/tex]

[tex]\mu = \dfrac{(9.8 \ m/s^2 ) sin (15^0) - \dfrac{ \dfrac{(25 \times 10^3}{3600} \ m/s)^2 }{100 \ m } cos (15^0) }{((\dfrac{(\dfrac{25 \times 10^3}{3600} )^2}{100}) sin 15^0 + (9.8 \ m/s^2) cos 15^0 ) }[/tex]

[tex]\mathbf{\mu = 0.216}[/tex]

Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both charges are doubled, and the distance between them is halved, the new force between them is

Answers

Answer:

The new force between them is increased by a factor of 16.

Explanation:

According to Coulombs law, the force of attraction between two (2) charges is given by the formula;

F = Kq1q2/r²

Given the following data;

q1 = 2q1

q2 = 2q2

r = r/2

Substituting into the equation, we have;

F = 2q1*2q2/(r/2)²

F = 4q1q2/r²/4

F = 4q1q2 * 4/r²

F = 16q1q2/r²

Therefore, the new force between them is increased by a factor of 16.

Determine the gravitational field of Earth at a height 2.88 x 10^8 m above its surface (the height of the moon above Earth). Earth's mass is 6.0 x 10^24 kg and its radius is 6.4 x 10^6 m.

Answers

Answer:

12

Explanation:

100 POINTS. PLEASE EXPLAIN

Answers

Answer:

Explanation:

Note the charge balls on the top and bottom row are identical. So those charges cancel each other out. The only charges in the net electric field are the two in the middle row.

Electric field strength = k*Q/r^2

= (8.99 *10^9) * (3-(-3)) * 5*10^(-6) / (2*0.5)^2

= 269700 N/C

Answer:

Explanation:

(b) cuz the 1st n 3nd row cancel out, net electric field will go from +3Q to -3Q. the direction is right.

A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?

Answers

Answer:

the answer is b luv .

Explanation:

For which medical procedure would Doppler ultrasound be most useful?

A.
Finding a lung tumor
B.
Fixing a pulled muscle
C.
Locating a broken bone in a finger
D.
Detecting a blockage in a heart artery

Answers

The answer is d.

Thank me by clicking the ❤️. Thanks!

Doppler ultrasound would be most useful in detecting a blockage in a heart artery.

What are the clinical uses of Doppler ultrasound?

By monitoring the rate of change in pitch, a Doppler ultrasound may calculate how quickly blood flows (frequency). A sonographer with training in ultrasound imaging applies pressure to your skin with a tiny, hand-held instrument (transducer) roughly the size of a bar of soap across the area of your body being scanned, moving from one place to another as required.

As an alternative to more invasive treatments like angiography, which involves injecting dye into the blood arteries to make them visible on X-ray images, this test may be performed.

Your doctor may use a Doppler ultrasound to assess for artery damage or to keep track of specific vein and artery therapies.

Learn more about  Doppler technology here:

https://brainly.com/question/6109735

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An athlete runs on track at a constant speed of 75 meters/min for 15 s. What is the total distance he covered during that time?

Answers

Answer:

45

Explanation:

1. Amy uses 30N of force to push a lawn mower 10 meters. How much work does she do?

Answers

Work=300J (basically multiplying)

While driving his sports car at 20.00 m/s down a four lane highway, a
man comes up behind a very slow school bus. The man is in a hurry so he
decides to pass the bus. If the man's car can accelerate at 6.00 m/s^2,
how long will it take for him to reach a speed of 35.00 m/s in order to pass
the bus? Round to the nearest hundredth. (2 decimal places) Type the
number only.

Answers

Answer:

2.50 s

Explanation:

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

Initial velocity (u) = 20 m/s

Acceleration (a) = 6 m/s²

Final velocity (v) = 35 m/s

Time (t) =?

Thus, we can obtain the time taken for the man to reach a speed of 35 m/s as follow:

v = u + at

35 = 20 + 6t

Collect like terms

35 – 20 = 6t

15 = 6t

Divide both side by 6

t = 15/6

t = 2.50 s

Thus, it will take the man 2.50 s to reach a speed of 35 m/s.

A toy car, mass of 0.025 kg, is traveling on a horizontal track with a velocity of 5 m/s. If
the track then starts to climb upwards, how high up the track can the car reach?

Answers

Answer:

1.25 m

Explanation:

This is the vertical height not the distance along the slope.

[tex]K=U\\\frac{1}{2}mv^{2} = mgh\\h = \frac{v^{2}}{2g}=\frac{25}{20}=1.25 m[/tex]

The height the car can reach if the the track starts to climb upwards is 1.2742 meters up.

What is kinetic and potential energy?

Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or its relation with its surrounding systems.

P.E. = mass × g × height

K.E. = 0.5 × mass × (velocity)²

Given that the toy car has a mass of 0.025 kg and is traveling on a horizontal track with a velocity of 5 m/s. Now, the car starts to climb up vertically, therefore, the kinetic energy will be converted to potential energy.

Kinetic Energy = Potential Energy

0.5 × mass × (velocity)² = mass × g × height

Cancel mass from both the sides,

0.5 × (velocity)² = g × height

0.5 × (5 m/s)² = 9.81 m/sec² × height

height = 1.2742 meters

Hence, the car will travel 1.2742 meters up.

Learn more about Kinetic and Potential Energy here:

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A car accelerates uniformly from rest and reaches a speed of 9.9 m/s in 11.4 s. The diameter of a tire is 86.9 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.

Answers

Answer:

Number of revolutions = 20.71 rev.

Explanation:

Given the following data;

Initial speed, u = 0m/s

Final speed, v = 9.9m/s

Time, t = 11.4secs

Diameter = 86.9cm to meters = 86.9/100 = 0.869m

To find the acceleration;

Acceleration, a = (v - u)/t

Acceleration, a = (9.9 - 0)/11.4

Acceleration, a = 9.9/11.4

Acceleration, a = 0.87m/s²

Now we would find the distance covered by the tire using the second equation of motion.

S = ut + ½at²

S = 0(11.4) + ½*0.87*11.4²

S = 0 + 0.435*129.96

S = 56.53m

The circumference of the tire is calculated using the formula;

Circumference = 3.142 * diameter

Circumference = 3.142 * 0.869

Circumference = 2.73m

Number of revolutions = distance/circumference

Number of revolutions = 56.53/2.73

Number of revolutions = 20.71 rev.

Therefore, the number of revolutions the tire makes during this motion is 20.71 revolutions.

A physics grad student has a machine that measures how
many electrons are whizzing through a ring. If she measures 2
Coulombs of charge flowing through the ring every second,
what is the current passing through the ring?
1 = _A

Answers

Answer:

2A

Explanation:

Given parameter:

Quantity of charge  = 2C

Time taken = 1s

Unknown:

The current passing through the ring = ?

Solution:

Current can also be defined as the quantity of charge that passes  through a conductor per unit of time:

         I  = [tex]\frac{q}{t}[/tex]  

 So;

          I  = [tex]\frac{2}{1}[/tex]   = 2A

g A box of mass 10 kg attached to a spring is pulled to a maximum of 44 cm. The box is released. What is the speed when the box reaches a point 10 cm from the equilibrium position

Answers

Answer:

The speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.

Explanation:

Mass attached to the spring, m = 10 kg

maximum displacement of the spring, A = 0.44 m

The spring constant is calculated from Hook's law;

F = kx

mg = kx

k = (mg) / x

k = (10 x 9.8) / 0.44

k = 222.73 N/m

The angular speed of the spring is calculated as;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega =\sqrt{\frac{222.73}{10} } \\\\\omega = 4.72 \ rad/s[/tex]

The speed when the box reaches a point 10 cm from the equilibrium position is calculated as;

[tex]v = \omega \sqrt{A^2-x^2} \\\\v = 4.72\sqrt{0.44^2-0.1^2}\\\\v = 2.02 \ m/s[/tex]

Therefore, the speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.

What is the wavelength of a wave that has a speed of 350 meters second and a frequency of 140 hert??

Answers

Wavelength is the distance between identical points (adjacent crests) in the adjacent cycles of a waveform signal propagated in space or along a wire. In wireless systems, this length is usually specified in meters (m), centimeters (cm) or millimeters (mm).

Hope this helped

PLEASE ANSWER-Why are loose electrons needed for heat conduction?

Answers

Answer:

Why do metals conduct heat so well? The electrons in metal are delocalised electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.

The Earth (geosphere) comprises several major layers, differing in chemical and mineral composition, material strength, and other physical properties. Which of the layers listed below constitutes the largest proportion of Earth's volume?
a) Lithosphere
b) Crust
c) Outer core ·
d) Inner core
e) Mantle

Answers

Answer:

Option E:

The mantle

Explanation:

The earth's mantle is the mushy, semi-solid portion of the earth that makes up most of the earth's volume. The mantle extends for a depth of about 2800km downwards into the earth, making it the largest internal portion of the earth.  It makes up about 84 percent of the earth's structure, leaving the core and the crust with 15 percent and 1 percent respectively.

Due its nature, convection currents are set up predominantly in the mantle of the earth, which leads to movements in the upper layers of the earth (the crust).

The mantle is large enough for the lighter crust to float on its surface.

A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a total contact area of 4 cm 2 with the ground. Calculate the pressure he exerts on the ground (include units), and write your answer to two significant figures.

Answers

Answer:

P = 1471500 [Pa]

Explanation:

We must remember that pressure is defined as the relationship between Force over the area.

[tex]P=F/A[/tex]

where:

P = pressure [Pa] (units of pascals)

F = force [N] (units of Newtons)

A = area of contact = 4 [cm²]

But first we must convert from cm² to m²

[tex]A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }[/tex]

A = 0.0004 [m²]

Also, the weight should be calculated as follows:

[tex]w = m*g[/tex]

where:

m = mass = 60 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

[tex]w = 60*9.81\\w = 588.6[N][/tex]

And the pressure:

[tex]P=588.6/0.0004\\P=1471500 [Pa][/tex]

Because 1 [Pa] = 1 [N/m²]

A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fast!!!)

Answers

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?

Answers

Answer:

A Impulse = – 25 Ns

B. Time = 5 s

Explanation:

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

Mass (M) = 5 Kg

Initial velocity (u) = 8 m/s

Final velocity (v) = 3 m/s

Impulse (I) =?

Time (t) =?

A. Determination of the Impulse.

Mass (M) = 5 Kg

Initial velocity (u) = 8 m/s

Final velocity (v) = 3 m/s

Impulse (I) =?

I = Ft = M(v – u)

I = M(v – u)

I = 5 (3 – 8)

I = 5 × – 5

I = – 25 Ns

NOTE: the negative sign indicates that the net force is acting in the negative direction.

B. Determination of the time.

Impulse (I) = 25 Ns

Force (F) = 5 N

Time (t) =?

I = Ft

25 = 5 × t

Divide both side by 5

t = 25 / 5

t = 5 s

Thus, it will take 5 s for the box to slide through the 15 m long ramp.

A Discuss the possibility of fracture of two leg bones that have a length of about 70cm and an average area of

about 4cm

2 when a 80kg person jump from a height of 300cm.

Noting: The breaking stress of the bone ϬB =1.5×108 N/m2 , and

Young’s modulus for the bone is Y=1.5×1010 N/m2

Answers

Answer: The bones won't fracture.

Explanation: Stress, in Physics, is a quantity describing forces that can cause deformation. Strain is the measure of how muc an object can be stretched or deformed. The ratio between stress and strain is called Young's modulus or elastic modulus

Breaking Stress of Bone is the maximum stress a bone can take before a rupture occur.

To determine if a person will break his/her bones by jumping from a height, we determine the energy necessary for that jump and compare it with the energy necessary to break a bone.

The energy for breaking a bone is calculated as

[tex]E=\frac{Al_{0}\sigma_{B}^{2}}{2Y}[/tex]

A is the area in m²

l₀ is length in m

[tex]\sigma_{B}[/tex] is breaking stress in N/m²

Y is Young's modulus in N/m²

Calculating energy to break a bone:

[tex]E=\frac{4.10^{-4}.7.10^{-1}.(1.5.10^{8})^{2}}{2.(1.5.10^{10})}[/tex]

[tex]E=210[/tex] J

This is the energy necessary to break one leg bone, so as there are 2, energy will be 420 Joules.

Potential energy gained by jumping is calculated as

E = m.g.h

m is mass in kg

g is acceleration due to gravity in m/s²

h is height in m

Calculating

E = 80.(9.8)(0.3)

E = 235.2 J

Comparing the two energies, potential energy for jumping is less than maximum energy a bone can absorve without breaking, so the leg bones won't suffer a fracture.

A car with mass 1141 kg, moving at 16 m/s, strikes a(n) 2916 kg car at rest. If the two cars stick together, with what speed do they move

Answers

Answer:

V = 4.49 m/s

Explanation:

Given that,

Mass of the car 1, m₁ = 1141 kg

Initial speed of car 1, u₁ = 16 m/s

Mass of car 2, m₂ = 2916 kg

Initial speed of car 2, u₂ = 0

We need to find the speed of the cars if they stick together. Let the speed be V. The momentum will remain conserved in the process. Using the conservation of momentum to find it.

m₁u₁ + m₂u₂ = (m₁+m₂)V

[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\=\dfrac{1141\times 16+2916 \times 0}{(1141 +2916 )}\\\\=4.49\ m/s[/tex]

So, the required speed of the two cars is 4.49 m/s.

3.
A person drives north 6 blocks, then drives west 6 blocks.
The displacement is a straight line from the starting point to the finish in a
direction
O A northwesterly
O B southwesterly
O c northeasterly
OD southeasterly

Answers

The answer is c northeasterly because it goes through the middle of those two

2. A girl of weight 400 N has feet of area 100 cm2. What

pressure does she put on the ground ?

Answers

Answer:

Explanation:

Given:

Weight = 400N

Area = 100cm^2

If 1cm^2 = 1×10^-4m^2

Hence 100 cm^2 = 0.01m^2

Pressure = force/area .........1

Force = ma = mg which is also the weight of the girl. Hence substitute the values into 1

P = 400/0.01

P = 40000 N/m^2

The pressure she puts on the ground is 40000 N/m^2

which hand has a negatively charged?

Answers

Answer:

The dryer sheet is negatively charged and your hand is positively charged

Explanation:

Help can’t find the answer no where

Answers

i'm stuck on that question also

A jumbo egg (80 grams) is dropped from a height of 15 meters onto a 1 inch of foam. Using kinematics, determine the velocity of the egg the instant before impact.

Answers

Answer:

the velocity of the egg the instant before impact is 17.15 m/s.

Explanation:

Given;

mass of the egg, m = 80 g = 0.08 kg

height through which the egg was dropped, h = 15 m

The velocity of the egg before impact will be maximum, and the final velocity is given by the following kinematic equation;

v² = u² + 2gh

where;

u is the initial velocity of the egg = 0

v is the final velocity of the egg before impact

v² = 0 + 2 x 9.8 x 15

v² = 294

v = √294

v = 17.15 m/s

Therefore, the velocity of the egg the instant before impact is 17.15 m/s.

Two force vectors are oriented such that the angle between their directions is 46 degrees and they have the same magnitude. If their magnitudes are 2.81 newtons, then what is the magnitude of their sum

Answers

Answer:

F = 5.17 N

Explanation:

If we know the magnitudes of both vectors, and the angle between them, we can find the magnitude of their sum, applying the cosine theorem, as follows:

       [tex]F =\sqrt{F_{1} ^{2} +F_{2} ^{2} + 2*F_{1} * F_{2} * cos \theta} (1)[/tex]

Replacing by the givens, F₁ = F₂ = 2.81 N, θ = 46º, we get:

       [tex]F =\sqrt{2.81 N ^{2} +2.81 N ^{2} + 2*2.81N* 2.81N* cos 46} = 5.17 N (2)[/tex]

Suppose the posted designated speed for a highway ramp is to be 30 mph and the radius of the curve is 700 ft. At what angle must the curve be banked? No Friction

Answers

Answer:

4.92°

Explanation:

The banking angle θ = tan⁻¹(v²/rg) where v = designated speed of ramp = 30 mph = 30 × 1609 m/3600 s = 13.41 m/s, r = radius of curve = 700 ft = 700 × 0.3048 m = 213.36 m and g = acceleration due to gravity = 9.8 m/s²

Substituting the variables into the equation, we have

θ = tan⁻¹(v²/rg)

= tan⁻¹((13.41 m/s)²/[213.36 m × 9.8 m/s²])

= tan⁻¹((179.8281 m²/s)²/[2090.928 m²/s²])

= tan⁻¹(0.086)

= 4.92°

Which statements correctly describe the formula or name of a compound? Select all that apply.
OA. The formula of nitrogen trifluoride is NF 3
B. The formula of ammonia is NH3.
C. The name of AlF, is trialuminum fluoride.
D. The formula of calcium chloride is CaCl2
E. The name of Li, Se is lithium selenate.
OF. The formula of dinitrogen monoxide is NO
2
Results
G. The formula of sulfur trioxide is 30.
Cho
Аа
H. The formula of magnesium hydroxide is Mg(OH)2
G.

Answers

Answer:

A, B, D, and H

Explanation:

Statements A, B, D and H are all correct except the following:

Statement C is incorrect. The name of [tex] AlF_3 [/tex] is aluminium fluoride NOT "trialuminium fluoride".

Statement E is incorrect. The name of [tex] Li_2Se [/tex] is lithium selenide NOT "lithium selenate".

Statement F is incorrect. Dinitrigen monoxide, also known as nitrous oxide has a formula of [tex] N_2O [/tex] NOT [tex] NO_2 [/tex].

Statement G is incorrect. Sulfur trioxide formula is [tex] SO_3 [/tex].

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