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ivanzaharov [21]
3 years ago
6

5.

Physics
1 answer:
igor_vitrenko [27]3 years ago
3 0

Answer:

a=40\ m/s^2

Explanation:

Given that,

Initial speed of a shuttlecock, u = 30 m/s

Final speed of the shuttlecock, v = 10 m/s

Time, t = 0.5 s

We need to find its average acceleration. The acceleration of an object is equal to the change in speed divided by time taken. It is given by :

a=\dfrac{v-u}{t}\\\\a=\dfrac{10-30}{0.5}\\\\a=-40\ m/s^2

So, the average acceleration of badminton shuttlecock is 40\ m/s^2.

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The meeting in orbit of two or more spacecraft is A. orbital countermand. B. orbital farce. C. orbital hit. D. orbital rendezvou
k0ka [10]
<span> D. orbital rendezvous</span>
6 0
3 years ago
One of the statements below is correct In an ideal system where all forms of friction are ignored?
noname [10]

Answer:

a.

Explanation:

In a perfect system then energy can be changed from Kinetic to Gravitational Potential and vice versa without any losses. A common form of loss would be thermal energy, so b. is not the answer. c. is invalid as we want these 2 forms of energy to be transferred between, and d. is invalid as these statements c and a contradict each other.

3 0
3 years ago
Pls help i will give u brainliest!
Ivan

Answer: B.

Explanation:

4 0
3 years ago
a vector points -43.0 units along the x-axis, and 11.1 units along the y-axis. find the magnitude of the vector
Sauron [17]

Answer:

the magnitude of the vector is 44,40 units

Explanation:

Using the Pythagoras theorem

M=\sqrt{(-43^{2} )+(11.1)^{2} } \\\\M= 44,4 units

3 0
3 years ago
Consider a supersonic missile flying at Mach 2.5 at an altitude of 10 km. Assume that the angle of the shock wave from the nose
NNADVOKAT [17]

Answer:

The distance at which nose of the vehicle will the shock wave impinge upon the ground is approximately 23km

Explanation:

Let

u be Mach angle

M be Mach Number = 2.5

h be altitude = 10km

d be distance between nose of the vehicle with which the shock wave impinge upon the ground

Using;

Sin u = 1/M

u = arcSin (1/M)

u = arcSin (1/2.5) = 23.58°

Using angle at a point rule = 90°

90 - 23.58 = 66.42°

Tan A = opp/adj

Tan 66.42° = d/10

2.2911 * 10 = d

d = 22.911km ≈ 23km

6 0
3 years ago
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