m
A 3.0 kg model train going right at 2.8 bumps into another 2.0 kg model train car moving in the same
S
m
direction at 1.6 . The heavier train car has a final speed of 2.2 to the right.
S
S
What is the final speed of the lighter 2.0 kg train car?

Answer:

Having a chronic disease

Explanation:

Answer:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Explanation:

Given

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]

Point: [tex](f(t0), g(t0), h(t0))[/tex]

[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information

Required

Determine the parametric equations

[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]

Differentiate with respect to t

[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]

Let t = 1 (i.e [tex]t0 = 1[/tex])

[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + k[/tex]

To solve for x, y and z, we make use of:

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]

This implies that:

[tex]r'(1)t = xi + yj + zk[/tex]

So, we have:

[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]

[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]

By comparison:

[tex]xi = it[/tex]

Divide by i

[tex]x = t[/tex]

[tex]yj = \frac{1}{3}jt[/tex]

Divide by j

[tex]y = \frac{1}{3}t[/tex]

[tex]zk = kt[/tex]

Divide by k

[tex]z = t[/tex]

Hence, the parametric equations are:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Answer:

which the substance is lost by catabolism and excretion.

Explanation:

Answer:

0.248

Explanation:

Initial speed u = 13.6

Final speed v = 0

Distance s = 38.1

We have umg = ma

We make u subject of the formula

u = a/g

V² = u² + 2as

a = v²-u²/2s

We substitute the values into the above

a = 0-(13.6)²/2*38.1

a = 184.96/76.2

a = 2.427m/sec

Remember that

u = a/g

u = 2.427/9.8

= 0.2476

This is approximately 0.248

This is the minimum coefficient of friction required to keep the crate from sliding.

Answer:

The speed of a cyclist is 0.3 km/min.

Explanation:

Given

To determine

We need to determine the speed of a cyclist.

In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.

We know the formula involving speed, time, and distance

[tex]s=\frac{d}{t}[/tex]

where

substitute d = 6, and t = 20 in the formula

[tex]s=\frac{d}{t}[/tex]

[tex]s=\frac{6}{20}[/tex]

Cancel the common factor 2

[tex]s=\frac{3}{10}[/tex]

[tex]s=0.3[/tex] km/min

Thus, the speed of a cyclist is 0.3 km/min.

Answer:

4.552m/s

Explanation:

[tex]V=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1} } =\frac{2400*4.33+60*8.88}{2400}=4.552m/s[/tex]

Explanation:

2forH2,1for02,and2forH20

Answer:

I believe you put how you think you'd feel it's that simple

Answer:

utilizing positive values

Explanation:

I think you need Graph to figure it out

Using Newton's second law and kinematic projectile motion we can find the proton deflection y = 5.43 10⁻⁷ m, in the opposite direction to the electron deflection.

given parameters

to find

For this exercise we will use Newton's second law where the force is electric

            F = ma

            F = q E

where F is the force, q the charge, E the electric field, m the mass and the acceleration of the particle

           q E = m a

           a = q / m E

This acceleration is the direction of the electric field that is perpendicular to the initial velocity (v_i)

Having the acceleration we can use the kinematics relations

If we make the direction of the initial velocity coincide with the x-axis

             v_i = cte

             v_i = x / t

             t = x/ v_i

on the y-axis is in the direction of the electric field

            y = v_{iy}  t + ½ a t²

on this axis the initial velocity is zero

            y = [tex]\frac{1}{2} (\frac{q}{m} E) \ t^2[/tex]

subtitute

            y =            (1)

Electron motion.

Let us propose the expression for the electron situation, the length of the displacement must be the same for electron and proton, suppose that it is x = L

In this case the charge q = -e and the mass m = m_e

its substitute in  equation 1

            y₁ = [tex]\frac{1}{2} \ ( \frac{-e}{m_e} E) \ \frac{x^2}{v_i^2}[/tex]  

where y₁, is the lectron deflection.

Proton motion

Between the proton and the electron we have some relationships

          q_p = -e

          m_ = 1840 m_e

we substitute in the equation  1

         y₂ = ½ e / 1840 me E x² / vi²

         y₂ =

         y₂ = - y₁ / 1840

         y₂ = - 0.001 / 1840

         y₂ = - 5.43 10⁻⁷ m

The negative sign indicates that the deflection of the proton is in the opposite direction to the deflection of the electron.

In conclusion they use Newton's second law and kinematics we can find the proton deflection is y = 5.43 10⁻⁷ m

learn more about electric charge movement here:  https://brainly.com/question/19315467

Answer:

All of then are true

I need brainliest so I can rank up

Explanation:

Answer:

I think all options are true is the right answer

Explanation:

Mark me the brainliest plzzz

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

[tex]V_{A}[/tex] = 1 m³

[tex]T_{A}[/tex] = 10°C = 283 K

[tex]P_{A}[/tex] = 350 kPa

[tex]m_{B}[/tex] = 3 kg

[tex]T_{B}[/tex] = 35°C = 308 K

[tex]P_{B}[/tex] = 150 kPa

Now, lets apply the ideal gas equation;

[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]

[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150

[tex]V_{B}[/tex] = 265.188 / 150  

[tex]V_{B}[/tex] = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, [tex]m_{A}[/tex] =  [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221

[tex]m_{A}[/tex]  = 4.309 kg

Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg

Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex]  = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

[tex]P_{f}[/tex] =  [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]

given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K

we substitute

[tex]P_{f}[/tex] =  ( 7.309 × 0.287 × 293)  / 2.77

[tex]P_{f}[/tex] =  614.6211119 / 2.77

[tex]P_{f}[/tex] =  221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

Answer:

Explanation:

 a )

Time for first ball to reach top position

v = u - gt

0 = vi - gt

t = vi / g

Time to reach balcony  while going downwards

= vi /g

Total time = 2 vi / g

Time to go down further to the ground = t₁

Total time = 2 vi / g + t₁

Time for the other ball to go to the ground = t₁

Time difference = ( 2 vi / g + t₁ ) - t₁

= 2vi / g .

( b )

v² = u² + 2gh

For both the throw ,

final displacement = h , initial velocity downwards = vi

( For the first ball also  , when it go down while passing the balcony , it acquires the same velocity vi but its direction is downwards.)

vf² = vi² + 2gh

vf = √ ( vi² + 2gh )

(c )

displacement of first ball after time t

s₁ = - vi t + 1/2 g t²  [ As initial velocity is upwards , vi is negative ]

displacement of second ball after time t

s₂ = vi t + 1/2 g t²

Difference = d =  s₂ - s₁

= vi t + 1/2 g t² - ( - vi t + 1/2 g t² )

d = 2 vi t .

Answer:

it would be downwards due to gravitational force

The gravitational force is determined as 7.37 x 10⁻⁷ N.

The gravitational force is determined by applying Newton's law of universal gravitation.

Mathematically, the formula for the Newton's law of universal gravitation is given as;

Fg = ( Gm₁m₂ ) / ( r² )

where;

Fg = ( 6.67 x 10⁻¹¹ x 63.7 x 9750.6 ) / ( 7.5² )

Fg = 7.37 x 10⁻⁷ N

Learn more about gravitational force here: https://brainly.com/question/72250

#SPJ1

Answer:

Explanation:

For original spring , compression in spring due to a load of 1355 kg is

x = 12 - 8.55 = 3.45 cm = .0345 m

spring constant = W / x

= 1355 x 9.8 / .0345

= 384898.55 N /m

Spring constant of new spring

k = 384898.55 - 5655 = 379243.55 N /m

New compression for new spring

= W / k

= 1355 x 9.8 / 379243.55

= .035 m

= 3.50 cm

Difference of compression = 3.50 - 3.45

= .05 cm .

In later case , car will be more lowered by .05 cm .

Answer:

2.5 m/s^2

Explanation:

First, convert 90 km/hr into m/s:

90/3.6 = 25 m/s

vf = final velocity = 25 m/s

vi = initial velocity = 0 m/s

t = time = 10 seconds

a = acceleration, unknown

Then, find a using the following equation:

(vf - vi)/t = a

(25 m/s)/10 s = 2.5 m/s^2

a = 2.5 m/s^2

Hope this helps!! :)

Solution :

Given that :

The energy of the protons, K.E. = 1.10 keV

                                                    [tex]$= 1.10 \times 10^3 \ eV $[/tex]

The current produced by the generator is I = 5 mA

                                                                        [tex]$= 5 \times 10^{-3} \ A$[/tex]

Now [tex]$1 \ eV = 1.6 \times 10^{-19 }\ J$[/tex]

Mass of the proton, m = [tex]$1.67 \times 10^_{-27} $[/tex] kg

Charge of the proton, [tex]$q_p = 1.6 \times 10^{-19} \ C$[/tex]

a). Therefore using the formula for K.E. we can find out the velocity of the proton.

[tex]$K.E. =\frac{1}{2}mv^2$[/tex]

[tex]$v=\sqrt{\frac{2K.E.}{m}}$[/tex]

[tex]$v=\sqrt{\frac{2\times 10^3 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}}$[/tex]

  [tex]$= 4.38 \times 10^5 \ m/s$[/tex]

b). We know that the current is :

 [tex]$I=\frac{\Delta Q}{\Delta t}$[/tex]

Therefore, the total charge in one second is given by :

[tex]$\Delta Q = I \times \Delta t$[/tex]

     [tex]$= 5 \times 10^{-3} \times 1$[/tex]

    [tex]$= 5 \times 10^{-3}\ C$[/tex]

So, the number of protons in this charge is given by :

[tex]$n = \frac{\Delta Q}{q_p}$[/tex]

  [tex]$=\frac{5 \times 10^{-3} }{1.6 \times 10^{-19}}$[/tex]

  [tex]$= 3.13 \times 10^{16}$[/tex] protons

Answer:

The correct option is c. 75 for this question

Explanation:

The correct option is c. 75 for this question:

Let's see how.

Continuity Equation is given as:

AcVc = AaVa

Where,

Aa = Area of Aorta

Ac = Area of the capillary

Va = Fluid speed in Aorta

Vc = Fluid speed in Capillary

So,

Assuming the fluid is the ideal one/

[tex]\pi[/tex]/4 [tex]Dc^{2}[/tex] Vc= [tex]\pi[/tex]/4 [tex]Da^{2}[/tex] Va

[tex]Dc^{2}[/tex] Vc= [tex]Da^{2}[/tex] Va

Dc = Da x [tex]\sqrt{\frac{Va}{Vc} }[/tex]

Dc = 2.5 cm x [tex]\sqrt{\frac{60 cm}{0.07 cm } }[/tex]

Dc = 73.192 cm

Dc = 75 approximately

Hence, the diameter of the capillary = 75 cm approximately  

Answer:I will say d

Explanation: because Potential energy is the energy stored within an object, due to the object's position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy.

Answer:

A)

The impulse that reaches the cell phone first is the Longitudinal wave

B)  0.0206  seconds

Explanation:

length of Iron rail = 7.5 m

speed of sound in Iron = 5950 m/s

speed of sound in Air = 343 m/s

A) Determine which pulse reaches the cell phone first

The impulse that reaches the cell phone first is the Longitudinal wave

 Time for longitudinal pulse to be detected =  7.5 / 5950 = 0.00126 s

  Time for pulse through air to be detected = 7.5 / 343 = 0.02186  s

B) separation in time between the arrivals of the two pulses

   ΔT = 0.02186 - 0.00126  = 0.0206  seconds

Answer:

d velocity will be the one according to me

An action which would help in preventing and coping with heat-related conditions is: A. Drinking water.

Heat can be defined as a form of energy that is transferred from a physical object (body) to another, as a result of a difference in temperature. Also, heat is a condition of weather that is generally characterized by a high degree of temperature.

This ultimately implies that, heat is most likely to cause dehydration and high body temperature.

In order to prevent and cope with heat-related conditions, you should ensure that you drink water at regular intervals for hydration.

Read more on heat here: brainly.com/question/12072129

Answer:

1. cell membrane

2. golgi body

3. mitochondrion

4. cytoplasm

5. nucleolus

6. nucleus

Explanation:

The correct answer to this question is Option A; 6.

Why?

In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.

~Thank you~