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zavuch27 [327]
3 years ago
13

7. DRAW A PICTURE TO SHOW WORK.

Physics
1 answer:
Degger [83]3 years ago
8 0
Answer:
Distance covered 28 km
displacement is 22.8 km North-East
Explanation;

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CCDs are _____ format detectors, which detect photons with _____ efficiency. analog, high analog, low digital, high digital, low
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7 0
3 years ago
A supersonic airplane is flying horizontally at a speed of 2610 km/h.
Delvig [45]

Answer:

Centripetal acceleration of this aircraft: approximately 6.52\; {\rm m \cdot s^{-2}}.

Distance covered during the turn: approximately 63.2\; {\rm km}.

Time required for the turn: approximately 0.0242\; \text{hours} (approximately 87.2\; {\rm s}.)

Explanation:

Convert velocity and radius to standard units:

\begin{aligned}v &= 2610\; {\rm km \cdot h^{-1}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &= 725\; {\rm m\cdot s^{-1}} \end{aligned}.

\begin{aligned} r = 80.5\; {\rm km} \times \frac{1000\; {\rm m}}{1\; {\rm km}} = 8.05 \times 10^{4}\; {\rm m}\end{aligned}.

Hence, the centripetal acceleration of this aircraft:

\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(725\; {\rm m\cdot s^{-1}})^{2} }{8.05 \times 10^{4}\; {\rm m}} \\ &\approx 6.53\; {\rm m \cdot s^{-2}}\end{aligned}.

The trajectory of the turn is an arc with a radius of r = 80.5\; {\rm km} and a central angle of \theta = 90^{\circ} = (\pi / 4). The length of this arc would be:

\begin{aligned} s &= r\, \theta \\ &= 80.5\; {\rm km} \times (\pi / 4) \\ &\approx 63.2\; {\rm km}\end{aligned}.

The time required to travel 63.2\; {\rm km} at a speed of 2610\; {\rm km \cdot h^{-1}} would be:

\begin{aligned}t &= \frac{s}{v} \\ &\approx \frac{63.2\; {\rm km}}{2610\; {\rm km \cdot h^{-1}}} \\ &\approx 0.0242\; {\rm h} \\ &\approx 0.0242 \; {\rm h} \times \frac{3600\; {\rm s}}{1\; {\rm h}} \\ &\approx 87.2\; {\rm s} \end{aligned}.

6 0
2 years ago
A rock of mass m = 0.0450 kg is attached to one end of a string and is whirled around in a horizontal circle. If the radius of t
AleksAgata [21]

Answer:

Tension, T = 0.1429 N

Explanation:

Given that,

Mass of the rock, m = 0.0450 kg

Radius of the circle, r = 0.580 m

Angular speed, \omega=2.34\ rad/s

The tension in the string is balanced by the centripetal force acting on it. It is given by :

T=\dfrac{mv^2}{r}

Since, v=r\omega

T=\dfrac{m(r\omega)^2}{r}

T=\dfrac{0.0450\times (0.580\times 2.34)^2}{0.580}

T = 0.1429 N

So, the tension in the string is 0.1429 N. Hence, this is the required solution.

8 0
3 years ago
ASAP!!!!!! I need help is my final
brilliants [131]

Answer:

bro the answer would be 15m/s north

Explanation:

                                   ;)

3 0
3 years ago
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