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3 years ago

PLEASE HELP ONLY 5 QUESTIONS

Physics

2 answers:

Phantasy [73]3 years ago

8 0

1. b or c
2. c
3. a? or d
4.
5. a

Butoxors [25]3 years ago

4 0

1. c

2. c

3. d

4. b

5. c

i just took the test and got them all right

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When do net force is applied to a moving object it still comes to rest because of its inertia

Paraphin [41]

That is false for that.

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3 years ago

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.

kow [346]

Before the engines fail, the rocket's altitude at time t 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 t 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 t 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.

3 0

3 years ago

Energy flows from the sun to _______ to consumers and eventually to _______

Kay [80]

Answer:

the answer is c. producers, detrivores

5 0

3 years ago

A spring with a spring constant value of 125 N/m is compressed 12.2 cm by pushing on it with a 215 g block. When the block is re

allsm [11]

Answer:

v = 2.94 m/s

Explanation:

When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.

Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.

Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means

(1/2)kx^2 = (1/2)mv^2

kx^2 = mv^2

v^2 = (kx^2)/m

v = sqrt((kx^2)/m)

v = x * sqrt(k/m)

v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg

v = 2.94 m/s

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2 years ago

a man throws a football straight into the air. As it rises,it slows down. Which type of energy is the football gaining?

kotykmax [81]

It is gaining potental energy which will then transfer to knetic energy as it falls

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3 years ago

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