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2 years ago

Find the expression for the displacement covered in nth or in last one second

Physics

1 answer:

Natali [406]2 years ago

3 0

Answer:

Snth = u + a/2 ( 2n - 1)

Explanation:

Do you need explanation based on graph, integration or other method?

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Student Exploration: Nuclear Decay. Has anyone done a Gizmos lab on this?

LUCKY_DIMON [66]

Answer:NOOPE

Explanation:IM THE MYSTERY MAN WHOOOOSHHHHHHHHHHHHHHH ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

6 0

2 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

A car is moving in uniform circular motion. If the cars speed were to double to keep the car moving with the same radius the acc

Stells [14]

Answer:

The centripetal acceleration would increase by a factor of 4

Correct choice: B.

Explanation:

Circular Motion

The circular motion is described when an object rotates about a fixed point called center. The distance from the object to the center is the radius. There are other magnitudes in the circular motion like the angular speed, tangent speed, and centripetal acceleration. The formulas are:

$v_t=w\ r$

$\displaystyle a_c=\frac{v_t^2}{r}$

If the speed is doubled and the radius is the same, then

$\displaystyle a_c=\frac{(2v_t)^2}{r}$

$\displaystyle a_c=4\frac{v_t^2}{r}$

The centripetal acceleration would increase by a factor of 4

Correct choice: B.

5 0

2 years ago

What is a newton equal to in terms of units of mass and acceleration

artcher [175]

1 newton is the force needed to accelerate 1 kilogram of mass
at the rate of 1 meter per second² .

1 N = 1 kg-m/s² .

It's a force equal to roughly 3.6 ounces.

7 0

2 years ago

Read 2 more answers

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

2 years ago

Find the expression for the displacement covered in nth or in last one second​

Find the expression for the displacement covered in nth or in last one second