- When an object is dropped what happens to potential and kinetic energy as it falls?

A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. a passenger standing on the platform

ipn [44]

The passenger is 2.3 m away from the door after 3 seconds

Since both train and passenger is moving with a constant acceleration so we can use the second equation of motion to calculate the distance cover by train and the passenger,

Distance cover by the train in 3 seconds

S=ut-0.5at^2

S=0*3-0.5*0.6*3^2=2.7 m

Distance cover by passenger in 3 seconds

S=0*3-0.5*1.2*3^2=5.4 m

Now distance of passenger from the train

s=5+2.7-5.4=2.3 m

Therefore after 3 seconds passenger is 2.3 m away from the door.

6 0

3 years ago

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I'm having trouble understanding the equation for average speed. Can someone please explain to me because I have a test tomorrow

serious [3.7K]

Average speed = total distance covered divided by the total time

7 0

3 years ago

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0

3 years ago

An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

d)

The angular displacement of the wheel in a certain time interval t is given by

$\theta=\omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

Two wires carrying equal currents exert a force ???????? on each other. (a) The current in each wire is doubled, while the dista

sweet-ann [11.9K]

a) The magnetic field created by a current-carrying wire is proportional to the current:

B ∝ I, B = magnetic field strength, I = current

The magnetic force acting on a current-carrying wire immersed in a magnetic field is proportional to the current and the magnetic field strength:

F ∝ IB, F = magnetic force, I = current, B = magnetic field strength

Let's focus on wire 1.

Since wire 2's current is doubled, wire 2 produces a magnetic field twice as strong as before.

Wire 1's current is also doubled, therefore we now have a wire having <em>twice as strong a current</em> immersed in <em>twice as strong a magnetic field</em>. The magnetic force on wire 1 (and you can make a similar argument for wire 2) will be four times as strong as before.

b) The general formula for the magnetic force acting on a current-carrying wire immersed in a magnetic field is given by:

F = IL×B

F = magnetic force vector

I = current

L = vector having a magnitude equal to wire length and representing direction of current

B = magnetic field vector

Note we are taking a cross product of the IL and B vectors, not the product of two scalar quantities.

The very nature of the cross product means that if L and B are parallel to each other, F = 0N

7 0

3 years ago