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MA_775_DIABLO [31]
3 years ago
8

Jayden was given a marshmallow and a syringe in class to experiment with. She placed the marshmallow in the syringe and sealed t

he end. When she depressed the plunger of the syringe what happened to the marshmallow and why?
Physics
2 answers:
densk [106]3 years ago
6 0

Answer:

The increased pressure, pushed the molecules closer together, and caused the marshmallow to shrink.

Explanation:

This follows Boyle’s law which states that pressure is inversely proportional to volume given that temperature is unchanged, (remains constant).

When Jayden presses the plunger, with a sealed end in order to stop the marshmallow from falling out, there is increased air pressure in the syringe, this increased pressure will push against the marshmallow causing it to shrink (reduce in volume). This justifies Boyle’s law, the syringe pressure is increased and consequently, the  marshmallow volume is decreased.

Len [333]3 years ago
3 0
The marshmallow shrunk because of the increased amount of air pressure. I know this because I have once decreased the air pressure, (put a marshmallow in vacuum chamber), and it expanded.
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A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re
marusya05 [52]

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

3 0
3 years ago
What occurs when two Stars collide into each other?
ddd [48]

Answer:

A stellar collision.

Explanation:

A stellar collision is the coming together of two stars caused by stellar dynamics within a star cluster, or by the orbital decay of a binary star due to stellar mass loss or gravitational radiation, or by other mechanisms not yet well understood.

5 0
3 years ago
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In concave lenses a Distance object appears ​
JulsSmile [24]

Answer:

it appears to be farther away than it actually is, and therefore smaller then the object itself.

4 0
2 years ago
This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

4 0
3 years ago
The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is
sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.

The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.

The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀

The distance between the two axes is L/2 (total length of the rod divided by 2

From the parallel axis theorem we have

I = I₀ + M(L/2)²

5 0
3 years ago
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