Fuzzy and kermit

A bird is standing on an electric transmission line carrying 3000 A of current. A wire like this has about 3.0 x 10-5 22 of resi

Darya [45]

Answer:

13.5 x 10^-9 A

Explanation:

Yes

6 0

3 years ago

An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the

ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, $E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J$

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.

The magnetic field in north direction, $B_y=19911.5\ nT$

The magnetic field in west direction, $B_x=-3257.1\ nT$

The magnetic field in vertical direction, $B_z=48381.8 \ nT$

Magnetic field, $B=(-3257.1i+19911.5j+48381.8k)\ nT$

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

$E_k=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E_k}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}$

$v=5.92\times 10^7 i\ m/s$ (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

$F=q(v\times B)$

$F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})$

Since, $i\times j=k\ \\j\times k=i\\k\times i=j$

And, $i\times i=j\times j=k\times k=0$

$F=1.6\times 10^{-19}\times [1178 k-2864.20j]$

$|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}$

$F=4.95\times 10^{-16}\ N$

(b) Let a is the acceleration of the electron. It can be calculated as :

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

$a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}$

$a=5.43\times 10^{14}\ m/s^2$

Hence, this is the required solution.

4 0

3 years ago

John and Daniel are playing tug-of-war together. John is exerting 10 N of force. Daniel is exerting 12 N of force. What is their

vovangra [49]

YEP ITS D 2 N in Daniel's direction

8 0

3 years ago

Read 2 more answers

The vector values are 4.0 km due East and 3.0 km due north the resultant vector is 5.0 km long and 37 north of East this values

klemol [59]

Answer:

This values shows a right angle triangle

Explanation:

Given;

a vector 4.0 km due East

a 3.0 km due north

the resultant vector is 5.0 km

The resultant vector can be obtained by Pythagoras theorem if the vectors form a right angle triangle.

R² = 4² + 3²

R² = 16 + 9

R² = 25

R = √25

R = 5 km (right angle triangle proved)

Therefore, this values shows a right angle triangle

4 0

3 years ago

How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?

jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

diameter of the steel, d = 4 mm

the radius of the wire, r = 2mm = 0.002 m

original length of the wire, L₁

final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)

extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁

the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

$A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2$

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}$

$F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N$

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0

3 years ago