How much time does it take a dropped object to fall 180 m on Earth?
18 s
36 S
10 s
6s

Answer:

please put pic of the questions

Answer:

Check Newton's Rings:

d = height of air film

s = distance from center to ring being considered

R = radius of circle considered

The approximate formula is:

d = s^2 / (2 R)   or s = (2 R d)^1/2

If we just use 4000 mi for R and 1/4 mi for d the height

we get s = (2 * 4000 * 1/4)^1/2 = 2000^1/2 mi = 45 mi

Answer:

2.66×10⁻⁹ N.

Explanation:

From the question,

Applying newton's law of universal gravitation,

Fg = GMm/r²............................... Equation 1

Where Fg = gravitational force, G = universal constant, M = mass of the mercury, m = mass of the human, r = radius of Mercury

Given: M = 3.31023 kg, M = 70 kg, r = 2.4106

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

Fg = 6.67×10⁻¹¹(70×3.31023)/(2.4106²)

Fg = 2.66×10⁻⁹ N.

Answer:

hello your question is incomplete attached below is the complete and the required circuit diagrams

answer :

Ai) This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

Explanation:

Ai) The voltage across the resistor will double when you double the current through the resistor

Given that : V = I*R.  

lets assume : I = 2 amperes , R = 3 ohms

V = 2*3 = 6 v

secondly lets assume double the value of  (I)   i.e. I = 4 amperes

hence : V = 4*3 = 12 volts

This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

Aii) Showing the two data points from simulation

I1                    V1            I2= 2I1         V2=2V1       V1/ I1 =V2/I2

0.9*10^3     9 * 10^3     1.8*10^3       18*10^3          10 ohms

1.6 * 10^3    16 * 10^3    3.2*10^3     32*10^3         10 ohms

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

Answer:

They are equal.

Explanation:

Answer:

a) 0, ±1.65  b)  ± 0.825m

Explanation:

This is a sound interference exercise, it can be described by the path difference between the two waves

for the case of constructive interference

          Δr = 2n λ/ 2                   n = 0, 1, 2 ...

for the case of destructive interference

          Δr = (2n + 1) λ/ 2

the speed of sound is related to the wavelength

          v = λ f

          λ = v / f

          λ= 340/206

          λ = 1.65 m

let's set a reference system in the center between the two speakers

a) let's find the distances for constructive interference

            Δr = 2n 1.65 / 2

            Δr = 1.65 n

* the first interference occurs at n = 0

          Δr = 0

therefore the interference in the center is maximum

* n = 1

          Dr = 1.65 m

the second inference occurs at 1.65 m from the center, therefore there is a right wing and a left wing,

We do not have any more interference between the speakers because

n = 2 Δr = 3.3m this distance can be from the speaker

b) let's look for the destructive interference points

           Δr = (2n + 1) 1.65 / 2

           Δr = (2n + 1) 0.825

m = 0       Δr = 0.825m

m = 1        Δr = 2,475m

We can see that we only have the first destructive interference, one on each side.

Answer:

f

Explanation:

f

Well we know it has to be greater than 300,000 km/s since we can't see it.

We can't calculate it any closer than that using the given information.

Answer:

Before providing an answer to the question, the values for acceleration given in questions A and B were written twice. So correction would go like this: For (a) when the truck accelerates at 2.20m/s2 northward, and for (b) when it accelerates at 3.40m/s2 southward.

The answer:

(a) 88N, northward.

(b) 78.4‬, southward.

Explanation:

(a) Maximum frictional force acting on the packing case= (coefficient of static friction) X (Normal force)

Normal force = mass X acceleration due to gravity

Maximum static frictional force acting on the packing case = (coefficient of static friction) X (mass of packing case  X acceleration due to gravity)

Maximum static frictional force = (0.30) X (40.0-kg) X (9.8m/s 2) = 117.6N

While Reaction force acting on the packing case = (mass of packing case) x (acceleration generated by the pickup truck)

Reaction force acting on the case = (40.0-kg) X (2.20m/s2) = 88N

With these values, one can conclude that the packing case is at rest since the reaction force of the case acting in the opposite direction is lesser than the frictional force. Making the magnitude and direction of the friction force acting on the case still move northward, and the static frictional force acting on equals the reaction force.

The answer is 88N, northward.

(b)  Here too, we need to still compare the reaction force with the value of the already determined Maximum static frictional force (117.6N) above. This is necessary to know the frictional force between the pickup truck"s floor and the packing case.

Reaction force acting on the case when acceleration is 3.40m/s2 = (40.0-kg) X (3.40m/s2) = 136‬ N

We can conclude that the reaction force (136‬ N) is greater than the maximum static frictional force (117.6N), suggesting that the packing case is in motion and the frictional force is no longer static.

This means a kinetic force is now acting on the pickup truck"s floor causing the packing case to also move. This kinetic force can be calculated as:

kinetic force = (coefficient of kinetic friction) X  (mass of packing case  X acceleration due to gravity)

= (0,20) X (40.0-kg) X (9.8m/s 2) = 78.4‬N

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

21560 J work must be done to raise an 1100 kg car 2m above the ground.

Work = mass * gravity * height

= 1100 * 9.8 * 2

= 21560 J

In precis, work is carried out whilst pressure acts upon an item to purpose a displacement. 3 portions must be regarded in the way to calculate the quantity of work. The ones 3 portions are force, displacement, and the perspective between the pressure and the displacement.

Paints carried out are elaborated in this type of manner that includes both forces exerted on the body and the whole displacement of the body. This block is preceded by a steady force F. The purpose of this pressure is to move the body a certain distance d in an immediate route in the route of the pressure.

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Answer:

a)  i = 0.746 A, b)   # _electron = 2.44 10¹⁹ electrons,  c)   E = 1.87 10⁴ J

Explanation:

a) The definition of current is the charge per unit of time

           i = Q / t

           i = 1.56 / 2.09

           i = 0.746 A

b) Let's look for the cargo in passing at this time

           i = Q / t

           Q = i t

            Q = 0.746 5.24

            Q = 3.904 C

an electron has a charge e = -1.6 10⁻¹⁹ C, let's use a direct proportions rule

            # _electron = 3.904 C (1 electron / 1.6 10⁻¹⁹)

            # _electron = 2.44 10¹⁹ electrons

the number of electrons has to be an integer

c) In this part you are asked to calculate the power

            P = V i

            P = 12 0.746

            P = 8.952 W

            P = E/t

            E = P t

            E = 8.952 2.09

            E = 1.87 10⁴ J

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, [tex]$V_{\infty}$[/tex] = 200 m/s

Given that the flow is laminar.

[tex]$Re_L=\frac{\rho V L}{\mu _{\infty}}$[/tex]

      [tex]$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$[/tex]

    [tex]$= 4.10 \times 10^7$[/tex]

So boundary layer thickness,

[tex]$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$[/tex]

[tex]$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$[/tex]

    = 0.0024 m

The dynamic pressure, [tex]$q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$[/tex]

                                           [tex]$ =\frac{1}{2} \times 1.225 \times 200^2$[/tex]

                                          [tex]$=2.45 \times 10^4 \ N/m^2$[/tex]

The skin friction drag co-efficient is given by

[tex]$C_f = \frac{1.328}{\sqrt{Re_L}}$[/tex]

     [tex]$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$[/tex]

     = 0.00021

[tex]$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$[/tex]

                  [tex]$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$[/tex]

                  = 270 N

Therefore the net drag = 270 x 2

                                      = 540 N

Answer:

A is the answer okkkkkkkkkkkkkkkk

Answer:

a protractor

Explanation:

because protractors measure angles

Answer:

Explanation:

The magnitude of the force is known as the resultant.

R = √Fx²+Fy²

Fx = 80cos 20 + 40cos70

Fx = 80(0.9397)+40(0.3420)

Fx = 75.176 + 13.68

Fx = 88.856N

Fy = 80sin 20 + 40sin70

Fy = 80(0.3420)+40(0.9397)

Fy = 27.36 + 37.588

Fy = 64.948N

R = √88.586²+64.948²

R = √7,847.48+4,218.24

R = √12,065.72

R = 109.5

R = 110N

Hence the magnitude of the forces is 110N

Answer:

Tangential speed or Rotational speed

Answer:

true

Explanation:

Answer:

true

Explanation:

it's up in Gogle trust me

Answer:

25 m/s

Explanation:

200/8 = 25

Answer:

c.o Kg • ms 08

Explanation:

put all together

The answer is option c) 0 kg m/s

It is a property of moving body by virtue of its mass and motion and that is equal to the product of the body's mass and velocity

momentum = mass * velocity

since , momentum is a vector quantity

in order to calculate momentum we need one of the quantity to be vector since , mass can never be vector as it is a scaler quantity we need to use velocity ( vector quantity ) not speed ( scaler quantity ) as multiplication of two scalers can never give rise to a vector quantity

so, the answer is option c) 0 kg m/s

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Answer:

The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.

Explanation:

Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.

Answer:

air

Explanation:

Answer:

b

Explanation:

because of bonds and the bowl is used for heating purpose

Answer:

I think b not sure but I hope its right and is that the nwea test??

Answer:

Compared to the first student, the second student did twice as much work as the first student.

Explanation:

The work done by the first student will be equal to the Force exerted by the backpack on the student carrying it multiplied by one mile (Distance).  The work done by the second student will be equal to the Force exerted by the backpack on the student carrying it multiplied by two miles (Distance).

Answer:

Circuit B

Explanation:

To know the correct answer to the question, we shall determine the reading of the ammeter i.e the current in each circuit. This can be obtained as follow:

For circuit A:

Resistance (R) = 5 ohms

Voltage (V) = 1.5 V

Current (I) =?

V = IR

1.5 = I × 5

Divide both side by 5

I = 1.5 / 5

I = 0.3 A

For circuit B:

We'll begin by calculating the total resistance in the circuit. This can be obtained as follow:

Resistance 1 (R₁) = 5 ohms

Resistance 2 (R₂) = 5 ohms

Total resistance (Rₜ) =?

Since they are in parallel connections, the total resistance can be obtained as illustrated below:

Rₜ = R₁ × R₂ / R₁ + R₂

Rₜ = 5 × 5 / 5 + 5

Rₜ = 25 / 10

Rₜ = 2.5 ohms

Finally, we shall determine the current in the circuit.

Resistance (R) = 2.5 ohms

Voltage (V) = 1.5 V

Current =?

V = IR

1.5 = I × 2.5

Divide both side by 2.5

I = 1.5 / 2.5

I = 0.6 A

For circuit C:

We'll begin by calculating the total resistance in the circuit. This can be obtained as follow:

Resistance 1 (R₁) = 5 ohms

Resistance 2 (R₂) = 5 ohms

Total resistance (Rₜ) =?

Since they are in series connections, the total resistance can be obtained as illustrated below:

Rₜ = R₁ + R₂

Rₜ = 5 + 5

Rₜ = 10 ohms

Finally, we shall determine the current in the circuit.

Resistance (R) = 10 ohms

Voltage (V) = 1.5 V

Current =?

V = IR

1.5 = I × 10

Divide both side by 10

I = 1.5 / 10

I = 0.15 A

SUMMARY:

Circuit >>>>>>> Current

A >>>>>>>>>>>> 0.3 A

B >>>>>>>>>>>> 0.6 A

C >>>>>>>>>>>> 0.15 A

from the above calculations, circuit B has the highest ammeter reading.

Answer:

E = 4156.02 Vm⁻¹

Explanation:

The magnitude of the uniform electric field between the plates can be given by the following formula:

[tex]E = \frac{\Delta V}{d}\\[/tex]

where,

E = Electric field strength = ?

ΔV = Potetial Difference = 293 V

d = distance between plates = 7.05 cm = 0.0705 m

Therefore,

[tex]E = \frac{293\ V}{0.0705\ m}\\\\[/tex]

E = 4156.02 Vm⁻¹

Answer:

The answer is "Choice C ".

Explanation:

The relationship between the E and V can be defined as follows:

[tex]\to E= -\Delta V[/tex]

Let,

[tex]\to E= \frac{\delta V}{\delta x}[/tex]

When E=0

[tex]\to \frac{\delta V}{\delta x}=0[/tex]

v is a constant value

Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.

Answer:

Thomas Edisons most famous invention was the phonograph

Thomas Edison announces his invention of the phonograph, a way to record and play back sound. Edison stumbled on one of his great inventions—the phonograph—while working on a way to record telephone communication at his laboratory in Menlo Park, New Jersey.

Explanation:

Hope I helped

Answer:

Explanation:

Suppose that the turbines of a coal-fired plant are driven by hot gases at a temperature of 886 K. the temperature of the exhaust area is only 305 K,  the efficiency of this heat engine

Answer:

Part A

Newton's second law of motion states that the force applied to an object is directly proportional to the rate of change of momentum that is produced

Mathematically, we have;

F = m·v - m·u/Δt

Where;

m = The mass of the object

v = The final velocity of the object

u = The initial velocity of the object

Δt = The duration of motion of the object during change in velocity

Therefore, given that the mass, 'M', of the truck is larger than the mass, 'm', of the minivan, where the time of change in velocity Δt, and the initial and final velocities of both automobiles are the same such as in a sudden stop, the garbage ruck will exert more force than the minivan, and therefore, the garbage truck has a greater initial momentum before the automobiles are brought to a stop

Part B;

According to Newton's first law of motion, we have;

The use of a seat belt (and airbag for front seated passengers) will prevent dashboard or windscreen for the front passengers or the front seat for the passengers in the back, from being the item that stops the continued forward motion of the passengers in the car, which can lead to injury

Part C; The Newton's law of motion that act on a body at the point of impact is Newton's third law of motion, which states that the action and reaction are equal and opposite

Therefore, the action of the garbage truck on the minivan upon impact is equal to the reaction of the minivan to the force the garbage truck exerts on the minivan

Explanation: