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Archy [21]
3 years ago
14

You can comfortably hold your fingers close beside a candle flame, but not very close above the flame. why? challenge (optional)

: why do candle flames quickly snuff out in gravity-free regions?
Physics
2 answers:
Inessa05 [86]3 years ago
6 0
The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.
vekshin13 years ago
3 0

Candle flame gases go up and not out, so your hand is okay with being next to the flame and not above it. You can also see the gases above the flame if you look. :)

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You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle
AlekseyPX

Answer:

v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

- mgh - F_f (s) = 0 - \frac{1}{2}mv^2

here we know that

F_f = \mu_k mg cos\theta

also we know that the length of the incline is given as

s = \frac{h}{sin\theta}

now we have

- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2

so we have

v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}

3 0
3 years ago
The set of frequencies of the electromagnetic waves emitted by the atoms of an element is called
VLD [36.1K]

Answer:

The set of frequencies of the electromagnetic Waves emitted by the atoms of an element is called emission spectrum.

7 0
3 years ago
A sailor pulls a crate across the deck of a ship with a rope, exerting a horizontal force of 150. N. The crate, which has a mass
nata0808 [166]

Answer:

B

Explanation:

7 0
3 years ago
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
The haystack maker tractor lifted 120 kg of grass to a height of 5 meters in 6 seconds. Calculate the power of the haystack make
Gelneren [198K]

Answer:

1000 Joules per second.

Explanation:

To lift that much grass 5 meters higher than it was before the tractor did work that is the product of the (force needed to overdo gravity) and (displacement):

W = mg\Delta h=120kg\cdot 10\frac{m}{s^2}\cdot 5m = 6000J

Power is work over the amount of time:

P = \frac{W}{\Delta t}=\frac{6000J}{6 s} = 1000 \frac{J}{s}

4 0
3 years ago
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