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Leto [7]
3 years ago
12

If there was no gravity acting on the object and it was launched at an upwards angle of 30 degrees, explain its path.

Physics
1 answer:
matrenka [14]3 years ago
4 0
C. Generally, gravity is the only force acting on a projectile (causing it to fall down towards the earth). If you took this away, it would have no forces acting on it and, according to Newton's 1st law of inertia, it would continue moving indefinitely
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Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ____________ not matching the type of st
rodikova [14]

Answer:

Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ___MODALITY_________ not matching the type of stimulus of the receptos

Explanation:

Because modality of sensation refers to what is perceived after the stimulus is effected eg the pressure modality is effected when the pressure receptors of the skin are stimulated so in this case sound will not elite olfactory sensation because the modality here is sound rather than smell

3 0
3 years ago
The ends of a magnet where thee forces are the strongest are called magnets
ladessa [460]
They are called magnets poles.
7 0
4 years ago
In electromagnetic waves, frequency is inversely proportional to what?
Illusion [34]
In ANY kind of wave, frequency is inversely proportional
to wavelength.

That means their product is always the same number ...
the wave speed.
3 0
4 years ago
Explain why an experiment should test only one variable at a time
padilas [110]
We use only one variable at a time to find the accurate result. We want to see how the result of experiment changes everytime with a single variable.
4 0
3 years ago
A 2 kg body is dropped from a height of 3 m. Calculate:
NikAS [45]

Answer:

   a) 6.26 m/s

   b) 7.67 m/s

Explanation:

The potential energy at height h0 is initially ...

  PE0 = mgh0

At height h1, the potential energy is ...

  PE1 = mgh1

The difference in potential energy is converted to kinetic energy:

  PE0 -PE1 = KE1 = (1/2)m(v1)^2

Solving for v1, we have ...

  mg(h0 -h1) = (1/2)m(v1)^2

  2g(h0 -h1) = (v1)^2

  v1 = √(2g(h0 -h1))

__

a) When the body is 1 m high, its speed is ...

  v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high

__

b) When the body is 0 m high, its speed is ...

  v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground

3 0
3 years ago
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