A metal bar moves through a magnetic field. the induced charges on the bar are

A rocket sled accelerates at 21.5 m/s^2 for 8.75 s. (a) What's its velocity at the end of that time? (b) How far has it traveled

Leno4ka [110]

Answer:

(a ) vf= 188.12m/s : Final speed at 8.75 s

(b) d= 823.04 m : Distance the rocket sled traveled

Explanation:

Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:

d =vi*t+1/2a*t² Formula (1)

vf= vi+at Formula(2)

Where:

vi: initial speed =0

a: acceleration=21.5 m/s²

t: time=8.75 s

vf: final speed in m/s

d:displacement in meters(m)

Calculation of displacement (d) and final speed (vf)

We replace data in formulas (1) and (2):

d= 0+1/2*21.5*8.75²

d= 823.04 m

vf= 0+21.5*8.75

vf= 188.12m/s

8 0

1 year ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current

sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero $U = 0.1355 \ m$

Explanation:

From the question we are told that

The distance of wire one from two along the y-axis is y = 0.340 m

The current on the first wire is $I_1 = (27.5i) A$

The force per unit length on each wire is $Z = 295 \mu N/m = 295*10^{-6} N/m$

Generally the force per unit length is mathematically represented as

$Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}$

=> $\frac{\mu_o I_1I_2}{2\pi y} = 295$

Where $\mu_o$ is the permeability of free space with a constant value of $\mu_o = 4\pi *10^{-7} \ N/A2$

substituting values

$\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}$

=> $I_2 = 18.23 \ A$

Let U denote the line in the xy-plane where the total magnetic field is zero

So

So the force per unit length of wire 2 from line U is equal to the force per unit length of wire 1 from line (y - U)

So

$\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }$

substituting values

$\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }$

$6.198 -18.23U = 27.5U$

$6.198=45.73U$

$U = 0.1355 \ m$

5 0

2 years ago

Can a small child play with fat child on the seesaw?Explain how?

RoseWind [281]

Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play

8 0

2 years ago

Read 2 more answers

A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g

kap26 [50]

I know you are Indian by your question, HC Verma class 9 or 11 !!

if you got any problem, comment !!