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Vlad [161]
4 years ago
12

An object moving with constant acceleration near the surface of the earth is influenced by

Physics
1 answer:
Ymorist [56]4 years ago
6 0
An object moving with constant acceleration near the surface of the earth is influenced by the gravity. <span>In physics, </span>gravitational acceleration<span> is the </span>acceleration<span> on an object caused by the force of </span>gravitation<span>. Neglecting friction such as air resistance, all small bodies accelerate in a </span>gravitational <span>field at the same rate relative to the center of mass.</span>
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How is the pressure exerted and the area on which it is exerted are related to each other?
andrezito [222]
Pressure = force/area
3 0
4 years ago
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
4 years ago
HELP ASAP PLZ!
Dafna1 [17]
1. is D. Scientific method cannot be replaced. 
2. is B. 
3. is B as well.
8 0
3 years ago
Read 2 more answers
The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin
alekssr [168]

Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = 200\mu m = 200\times 10^{- 6}\ m

Gauge Pressure inside, P_{in} = 25\ mmHg

Blood Pressure outside, P_{o} = 10\ mmHg

Now,

Change in pressure, \Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

\Delta P = \frac{4\pi T}{R}

T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

3 0
3 years ago
The figure shows the original channel of a river, as well as its current channel. Which of the following forces MOST LIKELY chan
mash [69]

Answer: bbbb

Explanation:

3 0
4 years ago
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