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3 years ago

Please help very easy 5th grade work giving brainliest

Physics

1 answer:

zhenek [66]3 years ago

8 0

Answer:

the answer is option B because opposit sides of the magnets attract each other

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Why does your weight change on the moon

lorasvet [3.4K]

Answer:

the moon has different gravitational force.

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3 years ago

Read 2 more answers

Steve tunes an old-fashioned radio that has an antenna made from a 3.0-cm-long solenoid with a cross-sectional area of 0.50 cm2

sukhopar [10]

Answer:

$L = 1.88\times10^-^4 H$

Explanation:

It is given that,

Length of the coil, l = 3 cm = 0.03 m

Area of cross section of the coil, $A=0.5\ cm^2=0.00005\ m^2$

Number of turns in the coil, N = 300

inductance of the coil is

$L=\dfrac{\mu_oN^2A}{l}$

$L=\dfrac{4\pi\times 10^{-7}\times (300)^2\times 5\times 10^{-5}}{0.03}\\L = 1.88\times10^-^4 H$

So, the inductance of the coil is 1.88× 10⁻⁴H

5 0

3 years ago

Is a dog barking an example of kinetic energy to potential energy

Pepsi [2]

It is not an example of kinetic to potential because the dog is already in motion rather then having the potential to do something. Hope this helps!

3 0

2 years ago

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.40 107 m/s and experiences an acceleration

noname [10]

Answer:

0.00956770833 T

Explanation:

q = Charge of electron = $1.6\times 10^{-19}\ C$

m = Mass of proton = $1.67\times 10^{-27}\ kg$

a = Acceleration = $2.2\times 10^{13}\ m/s^2$

v = Velocity = $2.4\times 10^{7}\ m/s$

Magnetic field is given by

$B=\dfrac{ma}{qv}\\\Rightarrow B=\dfrac{1.67\times 10^{-27}\times 2.2\times 10^{13}}{1.6\times 10^{-19}\times 2.4\times 10^{7}}\\\Rightarrow B=0.00956770833\ T$

The magnetic field is 0.00956770833 T

4 0

3 years ago

calculate minimum number of incident photons per area and of the minimum dose needed to visualize an object of 1 mm squared usin

Lynna [10]

Answer:

minimum number of photon is 4.05 × $10^{7}$

Explanation:

given data

energy = 50 keV = 50 × $10^{3}$ eV = 50 × $10^{3}$ × 1.602× $10^{-19}$ J

thickness = 10^-3

contrast = 1%

to find out

number of incident photons

solution

we know here equation that is

E = n × h × ν .......................1

put here all these value

50 × $10^{3}$ = n × 6.6× $10^{-34}$ × c/ 1× $10^{-3}$

50 × $10^{3}$ × 1.602× $10^{-19}$ = n × 6.6× $10^{-34}$ ×( 3 × $10^{8}$ / 1× $10^{-3}$ )

solve it and find n

n = 4.05 × $10^{7}$

so here minimum number of photon is 4.05 × $10^{7}$

3 0

3 years ago