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3 years ago

What features of your design helped your rocket hit the target?

Physics

1 answer:

jonny [76]3 years ago

8 0

Answer:

What features of your design helped your rocket hit the target? (Key factors include the rocket's weight, launch angle, ability to fly straight, and the balloon's pressure.) After testing, what changes did you make to your rocket and launcher? (Answers will vary.)

Explanation:

PLEASE GIVE ME BRAINLIEST IF NEW YOULL SEE NEAR MY ANSWER MARK AS BRAINLIEST IF NOT YOULL HAVE TO WAIT A DAY AND BRAINLY WILL REMIND YOU

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Please help which one is correct

Natalija [7]

Answer:

option (i) is correct

Explanation:

as there is no air resistance, no force is acting on the object horizontally, but gravitational acceleration will obviously act, regardless of the air resistance... option (i) is correct

8 0

3 years ago

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

$\lambda =5.98\times 10^{-7} m$

7 0

3 years ago

Which wave, A or B, has higher energy?

eduard

Answer:

Wave A.

Explanation:

The energy of a wave is directly proportional to the square of the amplitude.

If a wave has higher amplitude, it will have more energy. On the other hand, a wave having lower amplitude, it will have less eenergy.

In this case, we need to tell which wave has higher energy. Hence, the correct option is A because it has a higher amplitude.

8 0

3 years ago

Read 2 more answers

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$