how do you do question 2?

Answer:

In a series circuit, how does the voltage supplied by the battery compare to the voltage on each load? The voltage of the battery is equal to the voltage of each load added together. ... The voltage across the two resistors must both have the same voltage of the battery.

Explanation:

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Answer:

The voltage of the battery is equal to the voltage of each load added together. The voltage across the two resistors must both have the same voltage of the battery.

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Answer:

only long-range force that affects all particles is the gravitational force.

Explanation:

In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.

The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.

Consequently the only long-range force that affects all particles is the gravitational force.

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, [tex]\omega_i[/tex] = 10 rad/s

final angular velocity, [tex]\omega_f[/tex] = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

[tex]a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2[/tex]

Therefore, the angular acceleration of the car is 1.5 rad/s²

Answer:

by using v = u + at equation we can find "a"

14 = 60 - 2.7a

2.7a = 60 - 14

2.7a = 46

decceleration = 17.03

Answer:

A...................................

The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.

The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.

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The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

a. The force of the wall on the car and the car on the wall are equal

b. The force of the wall on the car is greatest

c. The force of the car on the wall is greatest

d. There is not enough information to tell"

Solution :

It is given that the lamp glows brightly for a shorter period of time when the switch is closed on it the switch is put on. But after the some time the lamp goes off and it stops working.

This is because as soon as we on the switch, the current start flowing to the lamp which makes the filament of the lamp to glow, but due to some issue, the current stop flowing even when the switch is on and this stops the lamp from glowing and hence the lamp does not work.

Answer:

[tex]T=62.9*10^{-9}[/tex]

Explanation:

From the question we are told that:

Index of refraction of Rinestones [tex]\gamma_1 =1.5[/tex]

Index of refraction of silicon [tex]\gamma_2 =2.0[/tex]

Wavelength  [tex]\lambda=576nm=576*10^{-9}[/tex]

Let each layer have thickness T

Therefore

Total Thickness =2T

Generally the equation for  Constructive interference is mathematically given by

[tex]2T=(m+0.5)\frac{l\lambda}{\gamma_2}[/tex]

Where

 [tex]M=0[/tex]

 [tex]2T=(0+0.5)\frac{576*10^{-9}}{2*2.0}[/tex]

 [tex]T=62.9*10^{-9}[/tex]

Explanation:

1. if an object sjnks in one liquid and floats on another liquid it implies that the density of second liquid is greater than the density of first liquid

Answer:

The mass of the dog food added is 9.03 kg

Explanation:

Given;

mass of the shopping cart, m₁ = 14.5 kg

let the mass of the bag added = m₂

the force applied, F = 12 N

initial velocity of the cart-bag system, u = 0

distance traveled by the system, d = 2.29 m

time of motion of the system, t = 3.0 s

The acceleration of the system is calculated as;

[tex]d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2[/tex]

The total mass of the system (M) is calculated as follows;

F = Ma

M = F/a

M = (12)/(0.51)

M = 23.53 kg

The mass of the dog food added is calculated as;

m₂ = M - m₁

m₂ = 23.53 kg - 14.5 kg

m₂ = 9.03 kg

Assume R is measured in meters (m) and M in kilograms (kg). Then

R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²

so t ² is indeed proportional to R ²/(GM).

Answer:

A.

Explanation:Object b will get a negative charge .

Answer:

Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

Answer:

1=8 ohms 2=0.5 Amps

Explanation:

Explanation:

u=40

v=?

h=?

v²-u²=2gs

0²-40²=2×10×s

160=20s

s=160/20

=80m/s

total distance= upward distance ×downward distance

=80+80

=160m

total displacement=0 because u and v is the same.

Answer:

The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Explanation:

Final velocity v = 0

Initial velocity u = 40m/s

We know that,

Using equation of motion

[tex]v^{2} =u^{2} +2gh[/tex]

[tex]0-40^{2} =2[/tex] × [tex]10[/tex] × [tex]h[/tex]

The maximum height is:

[tex]h=80[/tex] [tex]m[/tex]

The  stone will reach at the top and will come down

Therefore, the total distance will be:

[tex]s=h_{1} +h_{2}[/tex]

[tex]s=80m-80m=160m[/tex]

The net displacement is:

[tex]D=h_{1} -h_{2}[/tex]

[tex]D=80m-80m=0[/tex]

Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

hope this helps.....

Answer:

Friction Opposes Motion of an Object.

Now

To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force

Net force = Pushing Force - Frictional Force

Recall

Net Force; F=Ma

Ma = P - Fr

Now the question asked for How Much force Must be applied to Maintain a Constant velocity.

In a Constant Velocity Motion... Acceleration do not change... Its Zero

So Putting this into the formula above

M(0) = P - Fr

0=P - Fr

Fr = P.

This means

That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force

Since Frictional Force; Fr =223N

The Applied Force(Pushing Force) Must be equal to 223N too.

Answer:

Cause you are their family and they need you. Do you wanna be someone who abandoned them or be the one they look to and dont dislike into adulthood. Also you signed onto this job no one said it was going to be easy but you making your life easier but harder for someone who is just a kid who still needs you.

Answer:

the positive charge is the period number

Explanation:

I might be wrong

Answer:

The positive charge is the group number.

Explanation:

Answer:

como assim qual a tarefa que educação física? se você me explicar melhor eu consigo te responder !!

Explanation:

Answer:

a.

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

b.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N

Explanation:

a. The formula for finding the force of gravity, F, acting object on an object is given as follows;

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

Where;

F = The force acting between the Earth and the object

G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M = The mass of the Earth = 5.97 × 10²⁴ kg

m = The mass of the object

r = The distance between the center of the Earth and the object

b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;

The given mass of the satellite, m = 1,300 kg

The distance between the center of the Earth and the center of the satellite,  r = The length of the radius of the Earth + The height of orbit of the satellite

The given height of orbit of the satellite, h = 200 km

∴ r = R + h = 6,378 km + 200 km = 6,578 m

Therefore, by plugging in the values, we get;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c. Solving the above equation gives;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton

Answer:

C.) Longitudinal

Answer: C or B

Explanation:

The EMF induced in the wire moving perpendicularly through a magnetic field is 0.025V. The correct option is C.

The EMF is the electro motive force which causes the current to induce in the object moving in the magnetic field.

Given is the length of wire L =15cm =0.15m, magnetic field strength B = 1.4T and velocity of wire V =0.12 m/s

EMF is related to the length of wire, magnetic field strength  and velocity of wire proportionally.

ε = B x L x V

Plug the values, we get

ε = 1.4 x 0.15 x 0.12

ε = 0.025 Volts

Thus, the correct option is C.

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Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

Answer: The equations in column A is matched with gas laws in column B as follows:

21. PV = nRT : (g) Ideal gas law

22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law

23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law

24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law

25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law

26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law

27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion

Explanation:

(A) Ideal gas law: It states that the product of pressure and volume is directly proportional to the product of number of moles and temperature.

So, PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant

T = temperature

So, [tex]P_{1}V_{1} = P_{2}V_{2}[/tex]

[tex]V \propto T\\\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\V_{1}T_{2} = V_{2}T_{1}[/tex]

So,  [tex]P_{1}T_{2} = P_{2}T_{1}[/tex]

So, [tex]V_{1}n_{2} = V_{2}n_{1}[/tex]

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\or, P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex]

[tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex]

Thus, we can conclude that equation in column A is matched with gas laws in column B as follows:

21. PV = nRT : (g) Ideal gas law

22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law

23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law

24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law

25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law

26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law

27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion

Answer:

Kinetic Energy

Explanation: