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4 years ago

How does wood and plastic relate to forces

Physics

1 answer:

WARRIOR [948]4 years ago

6 0

Answer:

There are six kinds of forces that act on objects when they come into contact with one another: Normal force, applied force, frictional force, tension force, spring force and resisting force. These forces make objects change their motion or movement , the act of going from one place to another.

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A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01 ???? 10

SIZIF [17.4K]

Answer:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

h = 0.2 m

$P_{0}$ = 1.01 x $10^{5}$ Pa

$P_{1} V_{1} = P_{2} V_{2}$

$\frac{V_{2} }{V_{1}} = \frac{P_{1} }{P_{2}}$

$P_{1}$ = $P_{0}$ + ρgh, ρ = 1000 kg/ $m^{3}$

$P_{1}$ = 1.01 x $10^{5}$ Pa + (1000 x 9.8 x 0.2) = 1,0296 x $10^{5}$ Pa

$P_{2}$ = $P_{0}$ = $10^{5}$ Pa

thus,

$\frac{V_{2} }{V_{1}} = 1,0296 x [tex]10^{5}$ / $10^{5}$ = 1.019

8 0

3 years ago

A small lead ball, attached to a 1.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

soldier1979 [14.2K]

Answer:

h = 57.6 m

Explanation:

First, we find the linear speed of the ball while in circular motion:

v = rω

where,

v = linear speed of ball = ?

r = radius of circle = length of rope = 1.75 m

ω = angular speed = (3 rev/s)(2π rad/1 rev) = 18.84 rad/s

Therefore,

v = (1.75 m)(18.84 rad/s)

v = 32.98 m/s

Now, we apply the 3rd equation of motion on the ball, when it breaks:

2gh = Vf² - Vi²

where,

g = - 9.8 m/s² (negative sign due to upward motion)

h = height covered = ?

Vf = Final Velocity = 0 m/s (since, the ball finally stops at highest point for a moment)

Vi = Initial Velocity = 32.98 m/s

Therefore,

2(- 9.8 m/s²)h = (0 m/s)² - (32.98 m/s)²

h = ( - 1088.12 m²/s²)/( - 19.6 m/s²)

h = 55.5 m

since, the ball was initially at a height of 2.1 m from ground. So, the total height from ground, will now become:

h = 55.5 m + 2.1 m

7 0

3 years ago

three letters (JET) are placed in front of a plane mirror the image formed is in what arrangement???

mario62 [17]

Answer:

TEJ as this is a thing you wont get

3 0

3 years ago

Write a hypothesis about how the number of half-lives affects the number of radioactive atoms. Use the "if . . . then . . . beca

Roman55 [17]

Answer:

Explanation:

Answer:

Explanation:

The half life is the time taken for half of a radioactive substance to disintegrate.

The shorter the half life, the larger the decay constant and the faster the decay process.

For a very large half life, it would take a very long time for the radioactive nuclide to decay to half.

With each half life reached, a new set of daughter cell is formed. Atoms that have short half life would decay rapidly. Every radionuclide has its own characteristic half-life.

If the number of half-lives increases, then the number of radioactive atoms decreases, because approximately half of the atoms' nuclei decay with each half-life. With this observation, we can hypothesise and conduct experiment to support the assertion that as the number of half-lives increases then the number of radioactive atoms decreases.

8 0

4 years ago

Read 2 more answers

A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is

iris [78.8K]

(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.

(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
$\Phi = BA \cos \theta$
where $\theta$ is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case $\theta=90^{\circ}$ and so the cosine is zero, therefore the net flux is zero.

5 0

3 years ago