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3 years ago

10

Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in

to other types but is not itself produced?

1 answer:

labwork [276]3 years ago

4 0

Kinetic Energy

How to calculate

Kinetic energy is given by K.E.

$\\ \sf\longmapsto K.E=\dfrac{1}{2}mv^2$

Where

m is denoted to mass
v is denoted to velocity

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

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Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3 years ago

How many items are present in the compound (NH4)2Cr2O7

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In 2Cr2O7 there’s 2 items; Cr and O

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3 years ago

The billiard ball model was the working model of the atom until thomson discovered the

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<span>d. electron

J J Thomson discovered the electron, and it was put in his model of the atom.</span>

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3 years ago

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Some gamma ray bursts are hypothesized to come from mergers of neutron stars or black holes. if this hypothesis is correct, what

Umnica [9.8K]

We should see (and have now detected with LIGO) gravitational waves

6 0

2 years ago

Ivan

Answer:

car B will be 30 Km ahead of car A.

Explanation:

We'll begin by calculating the distance travelled by each car. This is illustrated below:

For car A:

Speed = 40 km/h

Time = 3 hours

Distance =?

Speed = distance / time

40 = distance / 3

Cross multiply

Distance = 40 × 3

Distance = 120 Km

For car B:

Speed = 50 km/h

Time = 3 hours

Distance =?

Speed = distance / time

50 = distance / 3

Cross multiply

Distance = 50 × 3

Distance = 150 Km

Finally, we shall determine the distance between car B an car A. This can be obtained as follow:

Distance travelled by car B (D₆) = 150 Km

Distance travelled by car A (Dₐ) = 120 Km

Distance apart =?

Distance apart = D₆ – Dₐ

Distance apart = 150 – 120

Distance apart = 30 Km

Therefore, car B will be 30 Km ahead of car A.

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3 years ago