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3 years ago

Changing the direction of current flow, will or will not affect the strength of an electromagnet? Pick one: Will or Will Not

Physics

2 answers:

Otrada [13]3 years ago

8 0

Answer:

Will not.

Explanation:

just got the question correct.

Genrish500 [490]3 years ago

4 0

Answer:

will not .

Explanation:

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if the flower pot in problem 3 falls off the windowsill and falls 20 meters downwards(i.e., is 10 meters from hitting the ground

morpeh [17]

Answer:

te energy is

Explanation:

8 0

3 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

6 0

3 years ago

Indicate whether the statement is true or false. The point located at the geometric center of an object is called the object's c

zmey [24]

This statement is true. The point located at the geometric center of an object is indeed called the object's center of gravity. Center of gravity is the average of the weight of a resultant of the parallel forces, including all the particles that is passing through its body.

4 0

3 years ago

A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy

AURORKA [14]

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

8 0

3 years ago

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of v0 = 15.0 m/s. The cliff i

ankoles [38]

Answer:

(a) x0 = 0m and y0 = 49.0m

(b) Vox = 15.0m/s Voy = 0m/s

(d) X = 15.0t and y = 49.0 - 4.9t²

(e) t = 3.16s

(f) Vf = 34.4m/s

Explanation:

8 0

3 years ago