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3 years ago

7

When heat is added to a substance, describe how the molecules are affected. Use the words kinetic energy and potential energy ap

propriately?

1 answer:

shepuryov [24]3 years ago

3 0

When heat is added to a substance, the molecules increase kinetic energy. in other words, the particules speed up.

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An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal

Ugo [173]

Answer:

The value is $T_t = 2.5659 \ s$

Explanation:

From the we are told that

The initial speed of the object is $u = 8 \ m/s$

The greatest height it reached is $h = 15 \ m$

Generally from kinematic equation we have that

$v^2 = u^2 + 2gH$

At maximum height v = 0 m/s

So

$0^2 = 8^2 + 2 * - 9.8 * H$

=> $H = 3.27 \ m$

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as

$s = h - H$

=> $s = 15 - 3.27$

=> $s = 11.73 \ m$

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

$v = u + (-g) t$

At maximum height v = 0 m/s

$0 = 8 - 9.8t$

=> $t = 0.8163 \ s$

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

$h = ut_1 + \frac{1}{2} gt_1^2$

Here the initial velocity is 0 m/s given that its the velocity at maximum height

Also g is positive because we are moving in the direction of gravity

So

$15 = 0* t + 4.9 t^2$

=> $t_1 = 1.7496$

Generally the total time taken is mathematically represented as

$T_t = t + t_1$

=> $T_t = 0.8163 + 1.7496$

=> $T_t = 2.5659 \ s$

6 0

3 years ago

Given that the atomic weight of hydrogen is approximately 1 gram per mole, use the method above to estimate the average molecula

denpristay [2]

Answer:

2

Explanation:

6 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

If a magnet is broken into two pieces, what happens to the magnetic poles?

Nikitich [7]

Answer:

Each piece will have a north pole and a south pole

Explanation:

5 0

3 years ago

Read 2 more answers

Use the following half-life graph to answer the following question:

Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

$m_0 = 50.0 g$

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

$m'=\frac{50.0 g}{2}=25.0 g$

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0

3 years ago