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mestny [16]
3 years ago
6

A balloon contains helium with a mass of 0.00296 g. What is the volume of helium in the balloon? (Note: Helium’s density is 0.00

001663 g/cm3.) Volume: 0.00296g/0.00001663g/cm^3 = 177.991581cm^3
Physics
1 answer:
Over [174]3 years ago
7 0

Answer:

178 cm3

Explanation:

From definition of density, it is mass per unit volume of an object, expressed as density=mass/volume and making volume the subject of the above formula we have volume= mass/density and substituting 0.00296 g for mass and 0.00001663 g/cm3 for density then we have

Volume=0.00296/0.00001663

Volume is approximately 178 cm3

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The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that both leg bones, togethe
wolverine [178]

Answer:

0.21m

Explanation:

Note that the 200 J before breaking legs is Kinetic energy

So if potential energy= kinetic energy

Then 200J = mgh

h= 200/75x 9.8

= 0.21m so the maximum height will be 0.21m

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3 years ago
Temperature is referred to as what Under the metric system
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4 0
3 years ago
What is an ampere?<br><br> NO GUESSING! ANSWERS ONLY!
Igoryamba

Answer:

the SI base unit of electrical current

7 0
2 years ago
Read 2 more answers
On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per
Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

V_{avg} = \frac{Final \ position\  - \ Initial \ position}{final \ time\  - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\  - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr

Therefore, the average velocity is 180 km/hr

3 0
3 years ago
the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located
vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

3 0
3 years ago
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