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3 years ago

How does the horizontal velocity of a projectile change throughout its motion?

1 answer:

3 0

The force of gravity pulls vertically down on the object. It does not pull left or right, so there are no horizontal forces due to gravity. This means the horizontal velocity remains constant throughout the object's projectile parabolic path.

Note: this is assuming wind resistance is ignored. If you account for wind resistance, then the object will slow down because the air particles push against the object and impede its progress.

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The student quickly discovers that placing the marshmallow over the flame is more effective than heating the marshmallow on the

ANEK [815]

Answer:

The answer is convection.

Explanation:

There are three types of heat transfer: conduction, convection and radiation.

Conduction occurs when two objects touch each other and transfer heat.

Convection occurs when an object heats its surrounding fluid (like air, or water) and, since the hot fluids are less dense than the cold ones, they go up. So convection is a type of heat transfer that usually goes from down to up.

Radiation occurs when objects emanate heat in the form of electromagnetic waves that propagates in all directions.

So in this case, when the marshmallow is above the fire, it is exposed to convection, which does not occur when it is on the side of the fire.

4 0

4 years ago

Sherlock Holmes examines a clue by holding his magnifying glass (with

Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm$

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

$M=-\frac{q}{p}=-\frac{-30}{10}=3$

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0

4 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Sloan [31]

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

4 0

4 years ago

If you dropped a rock and it fell 4.9 m in 1 s , how far would it fall in 3 s

GalinKa [24]

The formula relevant for this is:

h = v0t + 0.5 gt^2

since the rock was dropped, therefore:

h = 0.5 gt^2

we can see that:

h / t^2 = 0.5 g = constant

therefore:

4.9 m / (1 s)^2 = h / (3 s)^2

h = 44.1 m

6 0

4 years ago

An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A

melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

$\omega = \frac{\theta}{t}$

Where θ: is the angular displacement = π (since it moves halfway through a complete circle)

We have:

$t = \frac{\theta}{\omega} = \frac{\theta}{v/r}$

Where:

v: is the tangential speed

r: is the radius

The radius can be found equaling the magnetic force with the centripetal force:

$qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}$

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

$t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s$

Therefore, the time that takes for an alpha particle to move halfway through a complete circle is 4.21x10⁻⁷ s.

I hope it helps you!

4 0

3 years ago