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Law Incorporation [45]

3 years ago

14

What is the magnetic force of a 30m long power line that carries a current of 50A. The magnetic field runs perpendicular to the

power line and has a strength of 30T.
A. 45000 T/N

B. 45 V

C. 45 N

D. 45000 N

1 answer:

Ne4ueva [31]3 years ago

8 0

Answer:

F=BILsin90 when perpendicular sin90 =1 30T x50x30 so you can get 45000N

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A.Substitute in the units for each one and combine like terms.

storchak [24]

IDK ghjfnhgfjmrmhjgfhgfmmfh

8 0

3 years ago

7. A car moving at 10m/s (about 22.4 mph) crashes into a barrier and stops in 0.25 m.

Galina-37 [17]

Answer:

a) 0.05s

b) 4000N

Explanation:

a)When car is stopped its final velocity become zero

U- 10 m/s

V- 0 m/s

S - 0.25 m

t -?

S = (v+u)*t/2

0.25 =(10+0)*t/2

t = 0.05s

b) If we happened to calculate the avarage force we have to consider about acceleration

V= 0

U = 10

t = 0.05 s

a =?

V = U + at

0 = 10 -a * 0.05

a = 200 m/s2

F = m *a

= 20 * 200

= 4000N

6 0

3 years ago

Will mark brainliest! Please help!

djverab [1.8K]

Answer:

30 seconds

Explanation:

A = A02^-(t/hl)

--> ln(A/A0) = -(t/hl)ln2

solving for hl,

hl = -t x ln2 /ln(A/A0)

= -(60 min)xln2/ln(50/200)

= 0.5 min or 30 seconds

3 0

3 years ago

A car jack with a mechanical advantage of 6 needs to produce an output force of 600 N to raise a car. What input force is requir

Mkey [24]

B.) 100 N is the CORRECT answer

Please mark as brainliest! :)

6 0

4 years ago

Read 2 more answers

Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f

Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

$\lambda=380 nm =3.8\cdot 10^{-7}m$ (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

Substituting,

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J$

And keeping in mind that

$1 eV = 1.6\cdot 10^{-19}J$

This energy converted into electronvolts is

$E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV$

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0

3 years ago