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A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway worker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?

Answers

Answer 1

Answer:

Answer:22.76 m/s

Explanation:

Given

Train length(L)=75 m

Front of train after travelling 125 m is 18 m/s

Time taken by the front of train to cover 125 m

$v^2-u^2=2as$

$18^2-0=2* a* 125$

$a=1.296 m/s^2$

Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel a distance of 125+75=200 m

$v^2-u^2=2as$

$v^2=2* 1.296* 200$

$v=√(518.4)=22.76 m/s$

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What was the main idea of Malthus theory of population

Answers

Answer:

The idea that population growth is potentially exponential while the growth of the food supply or other resources is linear.

Explanation:

Name the four forces in physics?

Answers

Answer:

Gravitational

Electrostatic

magnetic

Frictional

gravitational

electrostatic

magnetic

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hope it helps

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The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?

Answers

The particle reach its minimum velocity at time 1.06 sec.

The function is given as

x=5t^3-8t^2+12

Differentiating the above equation with respect to time, to obtain the velocity

dx/dt=v=15t^2-16t

For maximum and minimum values, put dx/dt=0

15t^2-16t=0

On solving the equation, t=0, 1.06

Therefore at time t=1.06 sec, the particle has the minimum value of velocity.

The particle reaches its minimum velocity at t = 0 s or t = 16/15 s

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = (v - u)/(t) } }$

$\large {\boxed {d = (v + u)/(2)~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

$x = ( 5t^3 - 8t^2 + 12) ~ m$

To find the velocity function, we will derive the position function above.

$v = (dx)/(dt)$

$v = 5(3)t^(3-1) - 8(2)t^(2-1)$

$v = ( 15t^2 - 16t ) ~ m/s$

Next to calculate the time to reach its minimum speed, then v = 0 m/s

$0 = ( 15t^2 - 16t )$

$0 = t( 15t - 16)$

$\large {\boxed {t = 0 ~s ~ or ~ t = 16/15 ~ s} }$

Learn more

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.

Answers

Answer:

λ = -47 nC / m

Explanation:

The missing question is as follows:

" The potential difference between the surface of a 2.2 cm -diameter power line and a point 1.9 m distant is 3.8 kV. What is the magnitude of the line charge density on the power line? Express your answer using two significant figures. "

Given:

- The Diameter of the power line D = 2.2 cm

- The distance between two ends of power line L = 1.9m

- The potential difference across two ends V = 3.8 KV

Find:

What is the magnitude of the line charge density on the power line?

Solution:

- The derivation of the line of charges for a length L oriented along any axis centered at origin and the potential difference between two ends is as follows:

V = 2*k*λ*Ln( D / L )

Where,

k : Coulomb's Constant = 8.99*10^9

λ : The line charge density

- Re-arrange and solve for λ:

λ = V / 2*k*Ln( D / L )

Plug in the values:

λ = 3800 / 2*8.99*10^9*Ln( 2.2 / 190 )

λ = -4.74022*10^-8 C / m

λ = -47 nC / m

Final answer:

Line charge density is the total charge distributed along the length of a wire, expressed in coulombs per meter. To calculate it, divide the total charge by the total length of the wire. Without specific numbers for charge and length, a numerical value can't be given.

Explanation:

To calculate the magnitude of the line charge density of a power line, you need to know the total charge (Q) distributed along the total length (L) of the wire. The line charge density (λ) is then defined as λ = Q/L. Unfortunately, without any specific numbers provided for these parameters, I can't provide a numerical answer.

Line charge density is a significant concept in electromagnetism and is measured in coulombs per meter (C/m).

Remember that the charge can be uniform or non-uniform along the length of the line.

For example, if a power line has a total charge of 0.02 C spread along its length of 50 m, it would have a line charge density of λ = Q/L = 0.02 C / 50 m = 0.0004 C/m

Learn more about Line Charge Density here:

brainly.com/question/34815993

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A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The snowball falls straight back down in to a 6 inches of snow. The snowball feels a deceleration of 100 m/s2 upon impact with the snow bank before coming to rest. (a) When does the snowball hit the top of the snow bank? (b) How far from the ground does the snowball come a rest?

Answers

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\n\Rightarrow 0=4.5-32.1* t\n\Rightarrow (-4.5)/(-32.1)=t\n\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow s=4.5* 0.14+(1)/(2)* -32.1* 0.14^2\n\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+(1)/(2)at^2\n\Rightarrow 4.315=0t+(1)/(2)* 32.1* t^2\n\Rightarrow t=\sqrt{(4.315* 2)/(32.1)}\n\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\n\Rightarrow v=0+32.1* 0.518\n\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\n\Rightarrow s=(v^2-u^2)/(2a)\n\Rightarrow s=(0^2-16.62^2)/(2* -100* 3.28)\n\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.