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2 years ago

100 POINTS!!!

Physics

1 answer:

nignag [31]2 years ago

6 0

Answer:

very low air pressure and that's not 100 points

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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.

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

You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds

german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

$f_{beat}=\frac{1}{T}$

Then the frequency for the previous period given (2sec) is

$f_{beat}=\frac{1}{2}$

$f_{beat} = 0.5Hz$

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

$f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz$

Hence option A is incorrect.

We can do this process for 254Hz as $f_1$ and 258 Hz for $f_2$ , then

$f_{beat} =|254Hz-258Hz|$

$f_{beat} = 4Hz$

Hence option B is incorrect.

We can also do this process for 255Hz as $f_1$ and 257 Hz for $f_2$ , then

$f_{beat} =|255Hz-257Hz|$

$f_{beat} = 2Hz$

Hence option C is incorrect.

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

$f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz$

Hence option D is incorrect.

We can also do this process for 255.75Hz as $f_1$ and 256.25 Hz for $f_2$ , then

$f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz$

Hence option E is incorrect.

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0

3 years ago

Can you help me with my homework please ? Science

geniusboy [140]

The formula for average speed is S=D/T

1. S=72m/37s

Divide

S= 1.94

Kira's average speed is 1.94m/s.

2. S=7.5km / 1.5h

S=5

Your average speed is 5km/h

3. S=1260km/3.5h

S=360

The airplanes average speed is 360km/h

7 0

2 years ago

Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of

navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0

2 years ago

A spherical drop of water carrying a charge of 42 pC has a potential of 620 V at its surface (with V = 0 at infinity). (a) What

iren [92.7K]

Answer:

0.0006091222 m

Explanation:

q = Charge = 42 pC

V = Voltage = 620 V

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric potential is given by (at r = R)

$V=\dfrac{q}{4\pi\epsilon R}\\\Rightarrow R=\dfrac{q}{4\pi\epsilon V}\\\Rightarrow R=\dfrac{42\times 10^{-12}}{4\pi\times 8.85\times 10^{-12}\times 620}\\\Rightarrow R=0.0006091222\ m$

The radius of the drop is 0.0006091222 m

3 0

3 years ago