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Allisa [31]
3 years ago
13

What is the usual time ice takes to evaporate above a Bunsen burner?

Physics
1 answer:
Step2247 [10]3 years ago
5 0
Depends on the volume but a normal ice cube takes less than 20 seconds
You might be interested in
Define the term energy density of a body under strain​
vodka [1.7K]

Answer:

Please mark as Brainliest!!

Explanation:

Strain energy is defined as the energy stored in a body due to deformation. The strain energy per unit volume is known as strain energy density and the area under the stress-strain curve towards the point of deformation. When the applied force is released, the whole system returns to its original shape.

8 0
3 years ago
if a bicyclist, with initial velocity of zero, steadily gain speed until reaching a final velocity of 26m/s, how far away did sh
Alja [10]

Answer:

<h2><em>34.46m</em></h2>

Explanation:

Using one of the equation of motion to solve the question. According to the equation v² =u²+2as where;

v is the final velocity of the  bicyclist = 26m/s

u is the initial velocity of the  bicyclist = 0m/s

a is the acceleration due to gravity = 9.81m/s

s is the distance covered during travel

Substitute the given parameters into the formula above to get the distance traveled

26² = 0² + 2(9.81)s

676 = 19.62s

Divide both sides  by 19.62

676/19.62 = 19.62s/19.62

s = 34.46m

<em>The distance traveled by the  bicyclist during the race is 34.46m</em>

5 0
3 years ago
A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a m
mariarad [96]

Answer:

Explanation:

KE = ½Iω²

ΚΕ = ½(mL²/3)ω²

ΚΕ = ½(0.63(0.82²)/3)4.2²

ΚΕ = 1.24541928

KE = 1.2 J

5 0
2 years ago
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0
3 years ago
A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee
makkiz [27]

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹  * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

3 0
3 years ago
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