What can scan tiny details inside the human body by using principles of electromagnetism.

Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?

hichkok12 [17]

Answer:

$t=5.36\times 10^{-6}\ s$

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

$t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s$

So, the required time is $5.36\times 10^{-6}\ s$

7 0

2 years ago

2 ways weak nuclear forces and gravity are alike <br><br> PLEASE HELP ):

shtirl [24]

They are attractive
They don’t depend on charge

7 0

2 years ago

Which object has the most gravitational potential energy?

Kipish [7]

Answer: An 8 kg book at a height of 3 m has the most gravitational potential energy.

Explanation:

Gravitational potential energy is the product of mass of object, height of object and gravitational field.

So, formula to calculate gravitational potential energy is as follows.

U = mgh

where,

m = mass of object

g = gravitational field = $9.81 m/s^{2}$

h = height of object

(A) m = 5 kg and h = 2m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 2 m\\= 98.1 J (1 J = kg m^{2}/s^{2})$

(B) m = 8 kg and h = 2 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 2 m\\= 156.96 J (1 J = kg m^{2}/s^{2})$

(C) m = 8 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 3 m\\= 235.44 J (1 J = kg m^{2}/s^{2})$

(D) m = 5 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 3 m\\= 147.15 J (1 J = kg m^{2}/s^{2})$

Thus, we can conclude that an 8 kg book at a height of 3 m has the most gravitational potential energy.

3 0

2 years ago

cylindrical container is to be constructed to be open at the top with a volume of 27π cubic meters using the least amount of mat

Llana [10]

Answer:

radius comes out to be 3 m

height of the cylinder comes out to be 3m

Explanation:

given

volume of cylinder = 27π m³

π r² h = 27π

r² h = 27.............(1)

surface area of cylinder open at the top

S = 2πrh + π r²

$S = 2\pi \dfrac{27}{r} + \pi r^2$

$\frac{\mathrm{d} s}{\mathrm{d} r}=\frac{\mathrm{d}}{\mathrm{d} r} (2\pi \dfrac{27}{r} + \pi r^2)$

$\frac{\mathrm{d} s}{\mathrm{d} r}=54\pi \dfrac{-1}{r^2}+2\pi r$

$\frac{\mathrm{d} s}{\mathrm{d} r}=0$

for least amount of material requirement.

$\dfrac{54\pi }{r^2} = 2\pi r\\r=3m$

hence radius comes out to be 3 m

for height put the value in the equation 1

so, height of the cylinder comes out to be 3m

3 0

2 years ago

Read 2 more answers

A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d

Softa [21]

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

6 0

3 years ago