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3 years ago

In your own words, what is an electrical current?

Physics

1 answer:

prisoha [69]3 years ago

4 0

An electrical current is a flow of electric charge

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A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration

Eduardwww [97]

my answer is 1276.7 because I subtracted 88.3 miles and seconds from 1365 MI

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3 years ago

8 POINTS AND MARK BRAINIEST:

lesya692 [45]

The answer is A i just took the test.

3 0

3 years ago

Read 2 more answers

What visible wavelengths of light are strongly reflected from a 390-nm-thick soap bubble?

DiKsa [7]

Answer:

So visible wavelength which is possible here is

416 nm and 693.3 nm

Explanation:

As we know that for normal incidence of light the path difference of the reflected ray is given as

$2\mu t + \frac{\lambda}{2} = \Delta x$

so here we can say that for maximum intensity condition we will have

$\Delta x = N\lambda$

so we have

$2\mu t + \frac{\lambda}{2} = N\lambda$

now for visible wavelength we have

for N = 1

$2\mu t = \frac{\lambda}{2}$

$\lambda = 4\mu t$

$\lambda = 4(\frac{4}{3})(390 nm)$

$\lambda = 2080 nm$

for N = 2

$\lambda = \frac{4\mu t}{3}$

$\lambda = \frac{4(\frac{4}{3})(390 nm)}{3}$

$\lambda = 693.3 nm$

for N = 3

$\lambda = \frac{4\mu t}{5}$

$\lambda = \frac{4(\frac{4}{3})(390 nm)}{5}$

$\lambda = 416 nm$

6 0

4 years ago

Kevin jump straight up in the air to a height of 1 m at the top of his jump he has the potential energy of 1000 J answer the fol

NeTakaya

Answer:

1000N,4.48m/s

Explanation:

Height(h) = 1m

P. E = 1000J

P. E = mass x acceleration due to gravity x height

Acceleration due to gravity = 10m/s

P. E = mgh

Where weight = mg

P. E = weight x height

Weight = P. E / height

Weight = 1000 / 1

Weight. 1000N

b) as he leaves the ground P. E = K. E

P. E = 1/2 mv2

P. E = 1000 , mass = weight / 10

Mass =1000/10 = 100kg

P. E = 1/2 mv2

1000 = 1/2 x 100 x v2

Multiply both sides by 2

1000 x 2 = 100 x v2

V2 = 2000/ 100

V2 =20

Square root both sides

V = square root of 20

V = 4.48m/s

I hope this was helpful, Please mark as brainliest

4 0

3 years ago

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

4 years ago