Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster

Answer:

Time = 5 minutes

Answer:

2minutes per mile

Explanation:

Not sure if it helps

Answer: Option D.

Are equally spaced at both electrodes and between them.

Explanation:

An electric field is the region that surrond electrically charged particle which may cause attraction or repelling of the particle due to the force that is exerted on it.

The electric field between two electrodes due to electrostatic forces exist in the air and there is an increase of external voltage is added to it. Electric field are uniform between electrodes which signify that the electric field lines are evenly spaced at both electrodes and between them.

Answer:

  the vaulter has a height of 6.112 meters

Explanation:

Base runner: no question content.

__

Pole vaulter:

Initial kinetic energy* is ...

  KE = (1/2)mv^2 = (1/2)(45 kg)(11 m/s)^2 = 2722.5 J

Above the bar, her kinetic energy is ...

  KE = (1/2)mv^2 = (1/2)(45 kg)(1.1 m/s)^2 = 27.225 J

Then the amount of kinetic energy converted to potential energy is ...

  2722.5 J -27.225 J = 2695.275 J

This corresponds to a change in height of ...

  PE = mgh

  2695.275 J = (45 kg)(9.8 m/s^2)h

  h = 2695.275/(45·9.8) m = 6.112 m

Her altitude above the bar is 6.112 meters.

_____

* Here, we use the traditional equations for kinetic and potential energy, which use "m" to represent mass. When we fill in the numbers, we attach units to the numbers in which "m" represents "meters". We trust you can sort this without confusion.

Answer:

Acceleration a = 3.75 m/s^2

Explanation:

Given that

Mass M1 = mass of the truck = 2300kg

Mass M2 = mass of the SUV = 2500kg

Force applied F = 18000N

From Newton second law,

F = ma

Where a = acceleration

We must consider the entire mass involved in the system. Therefore,

F = (M1 + M2)a

Make a the subject of formula

a = F/(M1 + M2)

Acceleration a = 18000/( 2300 + 2500)

Acceleration a = 18000/ 4800

Acceleration a = 3.75 m/s^2

Answer:

  411.6 N

Explanation:

The force is given by ...

  F = ma . . . . m represents mass

  F = (42 kg)(9.8 m/s^2) = 411.6 N . . . . m represents meters

The force on the object due to gravity is 411.6 newtons.

Answer:

The lost in kinetic energy is   [tex]KE_l = 125.5 \ J[/tex]

Explanation:

From the question we are told that

   The mass of the first skater is  [tex]m_1 = 30 \ kg[/tex]

    The speed of the first skater is  [tex]v_1 = 3 \ m/s[/tex]

     The mass of the second skater is [tex]m_2 = 35 \ kg[/tex]  

     The  speed of the second skater is  [tex]v_2 = 1 \ m/ s[/tex]

     The final speed of both skater are [tex]v_f = 1.9 m/s[/tex]

The initial kinetic energy of both skaters is mathematically represented as

     [tex]KE_i = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2[/tex]

substituting values  

    [tex]KE_i = \frac{1}{2} * 30 * 3^2 + \frac{1}{2} * 35 * 1^2[/tex]

     [tex]KE_i = 242.5 \ J[/tex]

The final kinetic energy of both skaters is mathematically represented as

       [tex]KE_f = \frac{1}{2} * (m_1 + m_2 ) v_f^2[/tex]

substituting values

      [tex]KE_f = \frac{1}{2} * (30 + 35 ) * 1.9^2[/tex]

      [tex]KE_f = 117 \ J[/tex]

The lost in kinetic energy is

      [tex]KE_l = 242.5 -117[/tex]

       [tex]KE_l = 125.5 \ J[/tex]

Answer:

distance covered is 4.5km or 1.25m

Explanation:

GIVEN DATA:

SPEED=v=3km/h

TIME=t=1.5h

TO FIND:

DISTANCE COVERED=d=?

SOLUTION:

As we know that

speed=distance covered /time taken

here we have to find distance

speed×time taken=distance covered

3km/h×1.5h=distance covered

4.5km=distance covered or in international system of unit the answer is 1.25m

H= P × t

1kW= 1000w

10 min = 600s

H= 1000×600=600,000J

=> 143.40kcal

Answer:

Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.

Explanation:

I just learned about this 2 weeks ago actually.

Answer:

0.3c

Explanation:

The speed of Enterprise relative to Risa is 0.8c

Relative speed of both ships as measured from Enterprise is 0.5c

therefore, relative speed of Astra to Enterprise is 0.8c - 0.5c = 0.3c

this is also the relative speed with which Astra approaches the planet Risa since Enterprise's speed was calculated relative to Risa.

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

[tex]F=\dfrac{mv^2}{R}[/tex], m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

[tex]F=\dfrac{(16)^2m}{60}=4.26m\ N[/tex]

Case 2.

R = 15 m v = 8 m/s

[tex]F=\dfrac{(8)^2m}{15}=4.26m\ N[/tex]

Case 3.

R = 30 m v = 4 m/s

[tex]F=\dfrac{(4)^2m}{30}=0.54m\ N[/tex]

Case 4.

R = 45 m v = 4 m/s

[tex]F=\dfrac{(4)^2m}{45}=0.36m\ N[/tex]

Case 5.

R = 30 m v = 16 m/s

[tex]F=\dfrac{(16)^2m}{30}=8.54m\ N[/tex]

Case 6.

R = 15 m v =12 m/s

[tex]F=\dfrac{(12)^2m}{15}=9.6m\ N[/tex]

Ranking from largest to smallest is given by :

F>E>A=B>C>D

Answer:

its false

Explanation:

because if an leaf floats down from a tree it is not considered an object for a free-fall

Answer:

C

Explanation:

A feather falls from one end of a tube to the other inside a vacuum.

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

Answer:

Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy.

I hoep that was helpful! Let me know:)

Explanation:

Answer:

Potential Energy is when the energy is kept in place the object isnt moving so it isnt creating Kinetic energy.

Answer:

[tex]\boxed{\mathrm{view \ explanation}}[/tex]

Explanation:

Potential energy is the energy stored in a body due to the body’s positioning or state.

Answer:

Explanation:

Stiffness of spring k equal to 4 lb/ft.

Unstretched length of the spring L is equal to 0.5 feet.  

Weight of the collar W is 15lb

Radius of curvature of curve guide is 1 feet

length of vertical rod is 1.5 feet

Initial speed of collar when released from rest at A is 0 feet per seconds  

use the energy conservation equation

 [tex]P_A +K_A=P_B+K_B[/tex]

Estimate the potential energy , component as position B as below

[tex]P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft[/tex]

Estimate the kinetic energy , component as position A as below

[tex]K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft[/tex]

Estimate the kinetic energy , component as position B as below

[tex]K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2[/tex]

Substitute 20.5lb- ft for [tex]P_A[/tex]

0.5lb-ft for [tex]P_B[/tex]

0lb -ft for [tex]K_A[/tex]

[tex]0.0777V_2^2 for K_B[/tex]

[tex]20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05[/tex]

= 16.05ft/sec

Answer:

a)y = 485 m ,  v = 220 m / s , b)  y = 2954.39 m , c)   t_total = 51 s ,

d) v = 240.59 m / s

Explanation:

a) We can use vertical launch ratios for this exercise

the speed of the rocket the run out the fuel is

        v = v₀ + a t

the rocket departs with initial velocity v₀ = 0

        v = a t

        v = 55 4

        v = 220 m / s

the height at this point is

        y = y₀ + v₀t + ½ a t²

        y = y₀ + 1/2 a t²

        y = 45 + ½ 55 4²

        y = 485 m

b) the maximum height of the rocket is when its speed is zero

for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m

        v² = v₀´² + 2 g (y-y₀´ )

         0 = v₀´² + 2 g (y-y₀´ )

         y = y₀´ + v₀´² / 2g

         y = 485 + 220 2/2 9.8

         y = 2954.39 m

c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)

let's calculate the rise time

           v = v₀´- g t

           v = 0

            t₂ = v₀´ / g

            t₂ = 220 / 9.8

            t₂ = 22.45 s

Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor

           y = y₀´´ + v₀´´ t - ½ g t²

           0 = y₀´´ - ½ g t²

          t = √ (2 y₀´´ / g)

          t = √ (2 2954.39 / 9.8)

          t = 24.55 s

the total flight time is

       t_total = t₁ + t₂ + t₃

       t_total = 4 + 22.45 + 24.55

        t_total = 51 s

d) the veloicda right now

       v = vo + g t

       v = 9.8 24.55

       v = 240.59 m / s

Answer:

E = 4000 V / m

U = 1.92*10^-18 J

C' = 4.71 pF

1.2 times greater with di-electric

Explanation:

Given:-

- The potential difference between plates, V = 12 V

- The area of each plate, A = 7.6 cm^2

- The separation between plates, d = 0.3 cm

- The charge of the proton. q = 1.6*10^-19 C

- The initial velocity of proton, vi = 0 m/s

Solution:-

- The electric field ( E ) between the parallel plates of the air-filled capacitor is determined from the applied potential difference by the battery on the two ends of the plates.

- The separation ( d ) between the two plates allows the charge to be stored and the Electric field between two charged plates would be:

                          E = V / d

                          E = 12 / 0.003

                          E = 4,000 V/m ... Answer

- The amount of electrostatic potential energy stored between the two plates is ( U ) defined by:

                         U = q*E*d

                         U = (1.6 x10^-19)*(4000)*(0.003)

                         U = 1.92*10^-18 J  ... Answer

- The electrostatic energy stored between plates is ( U ) when the proton moves from the positively charges plate to negative charged plate the energy is stored within the proton.

- A slab of di-electric material ( Teflon ) is placed between the two plates with thickness equal to the separation ( d ) and Area similar to the area of the plate ( A ).

- The capacitance of the charged plates would be ( C ):

                        C = k*ε*A / d

Where,

            k: the di-electric constant of material = 2.1

            ε: permittivity of free space = 8.85 × 10^-12

- The new capacitance ( C' ) is:

                      C' = 2.1*(8.85 × 10^-12) *( 7.6 / 100^2 ) / 0.003

                      C' = 4.71 pF

- The new total energy stored in the capacitor is defined as follows:

                     U' = 0.5*C'*V^2

                     U' = 0.5*(4.71*10^-12)*(12)^2

                     U' = 3.391 * 10^-10 J

- The increase in potential energy stored is by the amount of increase in capacitance due to di-electric material ( Teflon ). The di-electric constant "k" causes an increase in the potential energy stored before and after the insertion.

- Hence, the new potential energy ( U' ) is " k = 2.1 " times the potential energy stored in a capacitor without the di-electric.

Answer:

7 minutes

Explanation:

From the question, the flow of the river applies equally to both the hat and the canoe. The best way to approach this problem is to look at it from a perspective that follows the water, and ignores the bank.

Hence, the water and the hat are stationary, and the man paddles away from the hat and then paddles back.

Now, the hat is at rest and relative to the water and the canoe always travels at a speed of 4 m/s relative to the water.

We are told that it takes 7 minutes for the man to travel away from the hat.

Since he is travelling at a constant speed, it will therefore take same time of 7 minutes for him to return.

So, it will take 7 minutes for the man to row back upriver to reclaim his hat.

Answer:

vi = 14.610

Explanation:

initial velocity of the first object (vi) = 0 m/s, because it was dropped

distance (y) = -20 m

initial position (y0) = 0 m

acceleration due to gravity (g) = -9.8 m/s^2

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)t^2 + 0 + 0

-20 = -4.9t^2

4.081 = t^2

+√4.081 = t

t = 2.020

time of second object = 2.020 - 1 = 1.020

Now we can plug in the new time to solve for vi of the second object.

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)1.020^2 + vi*1.020 + 0

-20 = -5.098 + 1.020vi

14.902 = 1.020vi

vi = 14.610 m/s

Answer:

Explanation:

charge transferred to sphere = 12 - 8 = 4 nC

a )

The positive charge is decreased on glass rod , that means electrons must have entered the glass rod from the sphere which reduces its positive charge.

b )

charge transferred = 4 nC = 4 x 10⁻⁹ C

charge on one electron = 1.6 x 10⁻¹⁹ C

No of electrons transferred

= Total charge transferred / charge on one electron

= 4 x 10⁻⁹ / 1.6 x 10⁻¹⁹

= 2.5 x 10¹⁰ .

This question involves the concepts of charges and charged particles.

a) "Electrons" were transferred "from the sphere to the rod"

b) "2.5 x 10¹⁰" charged particles were transferred.

a)

Since the charge on the rod decreases by 4 nC. Therefore, negatively charged particles, that is electrons must have moved from the sphere to the rod. Because electrons are the mobile particles that move from one atom to the other.

b)

The number of charged particles transferred can be found using the following formula:

q = ne

where,

q = charge transferred = 4 nC = 4 x 10⁻⁹ C

n = no. of charged particles (electrons) transferred = ?

e = charge on one electron = 1.6 x 10⁻¹⁹ C

Therefore,

[tex]n=\frac{4\ x\ 10^{-9}\ C}{1.6\ x\ 10^{-19}\ C}[/tex]

n = 2.5 x 10¹⁰

Learn more about charged particles here:

https://brainly.com/question/1936492?referrer=searchResults

Answer:

q3 = 21.9 nC

Explanation:

By the Gauss theorem you have that the electric flux in a Gaussian surface is given by:

[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex]      (1)

ФE: electric flux = -218Nm^2/C

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/(Nm^2)

You can consider the spherical shell as a Gaussian surface. Then, the net charge inside the surface is:

[tex]Q=-18.0nC+38.0nC+q_3[/tex]     (2)

where charge q3 is unknown charge of the third object:

You replace the equation (2) into the equation (1), and you solve for q3:

[tex]\epsilon_0 \Phi_E=-18.0*10^{-9}C+38.0*10^{-9}C+q_3\\\\\epsilon_0 \Phi_E=20*10^{-9}C+q_3\\\\q_3=(8.85*10^{-12}C^2/(Nm^2))(-218Nm^2/C)-20*10^{-9}C\\\\q_3=2.19*10^{-9}C=21.9nC[/tex]

hence, the charge of the third object is 21.9 nC

Answer:

f' = 358442.3 Hz

Explanation:

This is a problem about the Doppler's effect. To find the perceived frequency of the observers you use the following formula:

[tex]f'=f(\frac{v}{v\pm v_s})[/tex]

v: speed of sound = 342 m/s

vs : speed of the source (jet) = 1220 km/h (1h/3600s)*(1000 m/ 1km) = 338.88 m/s  (it is convenient to convert the units of vs to m/s)

f: frequency emitted by the source = 3270 Hz

f': perceived frequency

Due to the jet is getting closer to the observers, the sing of the denominator in equation (1) is minus (-). Then, you replace the values of f, vs and v:

[tex]f'=(3270s^{-1})(\frac{342m/s}{342m/s-338.88m/s})=358442.3 Hz[/tex]

Hence, the perceived frequency by the observers is 358442.3 Hz

Answer:

[tex]f_o=359466.42Hz[/tex]

Explanation:

You can solve this problem using doppler effect formula. The Doppler effect is the phenomenon by which the frequency of the waves perceived by an observer varies when the emitting focus or the observer itself moves relative to each other

Doppler effect general case is given by:

[tex]f_o=f\frac{v\pm v_o}{v \mp v_s} \\\\Where:\\\\f=Actual\hspace{3}frequency\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source[/tex]

Now, you can use the following facts:

[tex]+v_o[/tex] Is used when the observer moves towards the source

[tex]-v_o[/tex] Is used when the observer moves away from the source

[tex]-v_s[/tex] Is used when the source moves towards the observer

[tex]+v_s[/tex] Is used when the source moves away from the observer

Since the source is alone in motion towards the observer, the formula is given by:

[tex]f_o=f\frac{v}{v-v_s} \\\\Where:\\\\v_s=1220km/h\approx 338.8888889m/s\\v=342m/s\\f=3270Hz[/tex]

Therefore:

[tex]f_o=(3270) \frac{342}{342-338.8888889} = 359466.42Hz[/tex]

Answer:

[tex]a)-3346.8\;N \\ b)-4937.04\;N-m[/tex]

Explanation:

a) - In Free-body diagram :

At point D, the free body diagram of a man :

[tex]D = Fn\\Fn=T+mg\\put\;values\;in\;it\\ .\;\;\;\;=300+(80)(9.81)=1084.8\;N[/tex]

[tex]Mg=200\times9.81=1962\;N[/tex]

[tex]\sum Fx=0\; where\; Ax=0[/tex]

[tex]\sum Fy=0\; where\; Ay-Mg-Fn-T=0[/tex]

Then, put the value in the equation.

[tex]Ay=3346.8\;N[/tex]

b)-

[tex]Ma=Mg(AE)+Fn(AD)+T=4937.04\;N-m[/tex]

Answer:

Q1: B.2 Q2: B.Waxing crescent Q3: A.Waxing Gibbous

Explanation:

Answer:

C

Explanation:

It happens when the a portion of the moon passes through the earth's shadow

Answer:

Explanation:

Velocity relative to earth = .99537 c

= .99537 x 3 x 10⁸ m /s

= 2.98611 x 10⁸ m /s

distance relative to earth = 40 km = 40 x 10³ m .

Time measured by scientist

= 40 x 10³ /  2.98611 x 10⁸

= 13.395 x 10⁻⁵ s

= 134 μs.

Answer:

[tex]= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m[/tex]

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

[tex]H_{max}=\frac{V^2\sin ^2 \theta}{2g}[/tex]

g = 9.8m/s2

[tex]= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m[/tex]

Therefore, highest point that the cannon ball reaches is 168.7m