1 airplane
travel due north at 300 km while another airplane travels Due South and 300 km are there speed the same or their velocities the same ​

Answers

Answer 1

Answer:

Explanation:

Speed is scalar and velocity is vector. Vector values imply direction as well as magnitude. Therefore, speed and velocity are not the same. The speeds of these 2 planes are the same at 300km/hr, but the velocity of the plane traveling north is +300km/hr while the velocity of the plane traveling south is -300km/hr if we define north as positive and south as negative.


Related Questions


A car starts from rest. If its acceleration is 1.5 m/s2 in 1.5 seconds, then calculate the
distance travelled by it.

Answers

Answer:

1.6875 m

Explanation:

Here, we are given that,

Initial velocity (u) = 0 (as it starts from rest)Acceleration (a) = 1.5 m/s²Time taken (t) = 1.5 s

Then,

Distance (s) = ?

By using the second equation of motion,

s = ut + ½at²

s is distanceu is initial velocityt is timea is acceleration

⇒ s = 0(1.5) + ½ × 1.5 × (1.5)²

⇒ s = ½ × 1.5 × 2.25

⇒ s = ½ × 1.5 × 2.25

⇒ s = ½ × 3.375

s = 1.6875 m

Distance travelled by it is 1.6875 m.

b) When the muscles connected to the crystalline lens contract fully, its focal length is 16.5000 cm. With this focal length, how far away must an object be to form sharply focused images on the retina? (Note: this distance is called the far point of vision.)
c) When the muscles connected to the crystalline lens relax, the focal length is 9.0000 cm. With this focal length, how close must an object be to form sharply focused images on the retina? (Note: this distance is called the near point of vision.)
d) As people age, the crystalline lens hardens (a condition called presbyopia or “old-age” eyes) and can only vary in focal length from 12 to 15.60 cm. Calculate range of vision (the new near point and far point) for this older eye.
e) Based on part d) why might an older person hold the newspaper at arm’s length to read it?

Answers

Answer:

I have to go to work and figure it out

Please help I need this done

Answers

The 3rd option if not sorry I think that is the answer

An amusement park ride whisks you vertically upward. You travel at a constant speed of 15 m/s during the entire ascent. You drop your phone 4.0 s after you (and your phone) begin your ascent from ground level.
a. How high above the ground is your phone when you drop it?
b. Find the maximum height above the ground reached by your phone.

Answers

Answer:

a. 60 m

b. 71.48 m

Explanation:

Below are the calculations:

a. The phone's height above the ground = Speed x Time

   The phone's height above the ground = 15 x 4 = 60 m

b. Speed when phone drops, u = 15 m/s

At maximum height, v = 0

Use below formula:

v² = u² -2gh

0 = 15² + 2 × 9.8 × h

h = 11.48 m

Total height = 60 + 11.48 = 71.48 m

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate would the car have to accelerate so that a quarter placed on the back wall would remain in place?

Answers

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

[tex]f_s = \mu_s N[/tex] --- (1)

where;

[tex]f_s =[/tex] static friction force

[tex]\mu_s =[/tex] coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

[tex]N = ma_{min}[/tex]

From equation (1)

[tex]f_s = \mu_s ma_{min}[/tex]

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

[tex]F = f_s[/tex]

[tex]mg = \mu_s ma_{min}[/tex]

[tex]a_{min} = \dfrac{mg }{ \mu_s m}[/tex]

[tex]a_{min} = \dfrac{g }{ \mu_s }[/tex]

where;

[tex]\mu_s = 0.383[/tex] and g = 9.8 m/s²

[tex]a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }[/tex]

[tex]\mathbf{a_{min}= 25.59 \ m/s^2}[/tex]

7. An electric train moving at 20km/hrs
. Accelerates to a speed of 30km/hrs. in
20 sec, find the distance travelled in meters during the period of
acceleration​

Answers

Answer

NB:

- speed, U is measure in m/s

- acceleration, a is measured in m/s²

-time t in seconds , s

Therefore conversation must be made

Speed U = 20km/hrs

=20km÷1hr

But 20km= 20×1000=20000m

1hr= 1×60min×60sec=3600s

U=20000÷3600=5.56m/s

a=30km/hrs

=30km÷1hr

But 30km=30×1000=30000

1hr=3600s

a=30000÷3600=8.33m/s²

From the equation of motion

S=Ut + ½ at².

Where s= distance

S = 5.56m/s × 20s + ½(8.33m/s²)(20s)²

S = 1777.3m

Are you aware of human rights violation happening in the community and explain​

Answers

Answer:

Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”

Kaseem is taking his bicycle for a ride. His bicycle is a system, and its main purpose is to provide transportation. What is the main input into this system? What is the desired output of this system?

Answers

His bicycle that operates on a system for transportation

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

Answers

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Explanation:

It is known that formula for path difference is as follows.

[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex]    ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.

[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]

Hence, path difference is as follows.

[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]

For lowest frequency, the value of n = 0.

[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]

[tex]\lambda = 4 \Delta L[/tex]

where,

[tex]\lambda[/tex] = wavelength

The relation between wavelength, speed and frequency is as follows.

[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]

where,

[tex]\nu[/tex] = speed

f = frequency

Substitute the values into above formula as follows.

[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

Pete is investigating the solubility of salt (NaCl) in water. He begins to add 50 grams of salt to 100 grams of
room temperature tap water in a beaker. After adding all of the salt and stirring for several minutes, Pete
notices a solid substance in the bottom of the beaker. Which statement best explains why there is a solid
substance in the bottom of the beaker?
A. The salt he is using is not soluble in water.
B. The salt is changing into a new substance that is not soluble in water,
C. The dissolving salt is causing impurities in the water to precipitate to the bottom
D. The water is saturated and the remaining salt precipitates to the bottom

Answers

Answer:

would the answer be c

Explanation: that what i think in my opian

Answer:

A

Explanation:

A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thrusters fail and the rocket continues to coast upward with insignificant air resistance. (a) What is the maximum height reached by the rocket

Answers

Answer:

The maximum height reached by the rocket is 486.53 m

Explanation:

Given;

initial velocity of the rocket, u = 0

acceleration of the rocket, a= 20 m/s²

duration of the rocket first motion, t = 4 s

The distance traveled by the rocket before its thrust failed

h₁ = ut + ¹/₂at²

h₁ = 0 + ¹/₂ x 20 x 4²

h₁ = 160 m

The second distance moved by the rocket is calculated as follows;

The velocity of the rocket before its thrust failed;

v = u + at

v = 0 +  20 x 4

v = 80 m/s

This becomes the initial velocity for the second stage

At maximum height, the final velocity = 0

[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]

The maximum height reached by the rocket = h₁ + h₂

                                                                          = 160 + 326.53

                                                                          = 486.53 m

16. The sum of kinetic energies in an object.
17. The essential device in power plants that convert mechanical
energy to electricity.
18. The device that converts electricity back to mechanical energy
19. The only EM wave that is seen by naked eye.
20. A device that converts light to electricity.​

Answers

Yes I also need help on this

which team won the champions league in 2020 2021​

Answers

Answer:

Chelsea F.C

Explanation:

Chelsea F.C

Soccer

Hai điện tích điểm Q1 = 8 C, Q2 = –6
C đặt tại hai điểm A, B cách nhau 0,1
m trong không khí. Tính cường độ điện
trường do hai điện tích này gây ra tại
điểm M, biết MA = 0,2 m

Answers

Answer:

English please

Explanation:

I don't understand the question

E=kq/r^2 chứng minh điện thế V=kq/r từ mối liên hệ giữa điện trường E và điện thế V

Answers

Answer:

hindi ko maintindihan teh

A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction

Answers

Answer:

The kinetic energy and potential energy lost to friction is 2,420 J.

Explanation:

Given;

total mass, m = 40 kg

initial velocity of the girl, Vi = 5 m/s

hight of the hill, h = 10 m

length of the hill, L = 100 m

initial kinetic energy of the girl at the top hill:

[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]

initial potential energy of the girl at the top hill:

[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]

Total energy at the top of the hill:

E = 500 J + 3920 J

E = 4,420 J

At the bottom of the hill:

final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s

hight of the hill = 0

final kinetic energy of the girl at the bottom of the hill:

[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]

final potential energy of the girl at the bottom of the hill:

[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]

Based on the principle of conservation of energy;  

the sum of the energy at the top hill = sum of the energy at the bottom hill

The energy at the bottom hill is less due to energy lost to friction.

[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]

Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.

True or False. A person who is nearsighted cannot see objects that are close to them clearly.

Answers

false, farsightedness is when you cant see close

Answer:

false

Explanation:

hope it works

A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires a speed of 1.7 m/s. Calculate the speed of the bullet before striking the block. Answer in units of m/s

Answers

Answer:

the speed of the bullet before striking the block is 302.3 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 28.3 g = 0.0283 kg

mass of the wooden block, m₂ = 5004 g = 5.004 kg

initial velocity of the block, u₂ = 0

final velocity of the bullet-wood system, v = 1.7 m/s

let the initial velocity of the bullet before striking the block = u₁

Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.

m₁u₁  +  m₂u₂  =  v(m₁  +  m₂)

0.0283u₁  + 5.004 x 0   =  1.7(0.0283  +  5.004)

0.0283u₁   =   8.5549

u₁ = 8.5549 / 0.0283

u₁ = 302.3 m/s

Therefore, the speed of the bullet before striking the block is 302.3 m/s.

How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?
Round off the answer to the nearest whole number.

Answers

Answer:

9.2 Relating Pressure, Volume,

Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Explanation:

hope its help :)

nicsfrom #philippines

A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brakes.

Answers

Answer:

V = 11.83 m/s

Explanation:

Given the following data;

Mass = 2000 kg

Force = 10000N

Distance = 14 m

To find the final velocity of the car;

First of all, we would determine the acceleration of the car;

Acceleration = force/mass

Acceleration = 10000/2000

Acceleration = 5 m/s²

Next, we would use the third equation of motion to find the final velocity;

[tex] V^{2} = U^{2} + 2aS [/tex]

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

V² = 0² + 2*5*14

V² = 0 + 140

V = √140

V = 11.83 m/s

Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5100 K. Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

[tex]V_{rms}=5.6*10^3m/s[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=5100K[/tex]

Generally the equation for RMS Speed is mathematically given by

 [tex]V_{rms}=\sqrt{\frac{3kT}{m}}[/tex]

Where

 [tex]K=Boltzman's constant[/tex]

 [tex]K=1.38*10^{-23}[/tex]

And

 [tex]M=molecular mass[/tex]

 [tex]M=4*1.67*10^{-27}[/tex]

 [tex]V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}[/tex]

 [tex]V_{rms}=5.6*10^3m/s[/tex]

Write down the chemical formula and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate

Answers

Answer: The chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.

Explanation:

The chemical formula of sodium bicarbonate is [tex]NaHCO_{3}[/tex].

In this compound, there are 1 sodium atom, 1 hydrogen atom, 1 carbon atom and three oxygen atoms present.

Therefore, the ratio of atoms is 1 : 1 : 1 : 3

Thus, we can conclude that the chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.

A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .

Required:
How far from the second lens is the final image of an object infinitely far from the first lens?

Answers

Answer:

the required distance is 6 cm

Explanation:

Given the data in the question;

f₁ = 15 cm

f₂ = 5.0 cm

d = 45 cm

Now, for first lens object distance s = ∝

1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'

Now, image distance of first lens s' = 15cm  

object distance of second lens s₂ will be;

s₂ = 45 - 15 = 30 cm

so

1/f₂ = 1/s₂ + 1/s'₂

1/5 = 1/30 + 1/s'₂

1/s'₂ = 1/5 - 1/30  

1/s'₂ = 1 / 6

s'₂ = 6 cm

Hence, the required distance is 6 cm

 

The distance of the final image from the first lens will be is 6 cm.

What is mirror equation?

The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).

The given data in the problem is;

f₁ is the focal length of lens 1= 15 cm

f₂ s the focal length of lens 2= 5.0 cm

d is the distance between the lenses = 45 cm

From the mirror equation;

[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]

If f₁ is the focal length of lens 1 is 15 cm then;

[tex]s'=15 cm[/tex]

f₂ s the focal length of lens 2= 5.0 cm

s₂ = 45 - 15 = 30 cm

From the mirror equation;

[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]

Hence the distance of the final image from the first lens will be is 6 cm.

To learn more about the mirror equation refer to the link;

https://brainly.com/question/3229491

Transfer of thermal energy between air molecules in closed room is an example of

conduction

convection

radiation


Answer and I will give you brainiliest ​

Answers

Answer: Conduction

Explanation: Conduction is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules. Conduction occurs more readily in solids and liquids, where the particles are closer to together, than in gases, where particles are further apart.

5. Tests performed on a 16.0 cm strip of the donated aorta reveal that it stretches 3.37 cm when a 1.80 N pull is exerted on it. (a) What is the force constant of this strip of aortal material

Answers

Answer:

53.41 N/m

Explanation:

From Hooke's law,

Applying,

F = ke............. Equation 1

Where F = Force, e = extension, k = force constant of the aortal material

Make k the subject of the equation

k = F/e............. Equation 2

From the question,

Given: F = 1.8 N, e = 3.37 cm = 0.0337 m

Substitute these values into equation 2

k = 1.8/(0.0337)

k = 53.41 N/m

Hence the force constant of the aortal material is 53.41 N/m

Define Potential Energy
Begin by defining potential energy in your own words within one concise eight word sentence

Answers

Answer:

potential energy is a type of energy an object has because of it's position

If you drive first at 40 km/h west and later at 60 km/h west, your average velocity is 50 km/h west.

Answers

and what else? is that all?

The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?

Answers

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N   average force applied

The magnitude of the average total force acting on the object during this time interval is 20 N.

The given parameters:

Mass of the object, m = 2.0 kgInitial velocity, u = 30 m/sFinal velocity, v = 40 m/sTime of motion, t = 5.0 s

The magnitude of the average total force acting on the object during this time interval is calculated as follows;

[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]

Learn more about resultant force here:  https://brainly.com/question/25239010

please help me and i will mark u as brainlist

Answers

Answer:

a) 8 secs I think

b)2m/s^2

I need help with physics question.

Answers

(D)

Explanation:

Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is

F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)

= 2.0×10^-11 N

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