A negative ion has more question 1 options:

Matter comprises all of them and among them are independent. what are they

telo118 [61]

Matter is composed of elementary particles i.e. quarks and leptons.

Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.

Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.Lepton is an elementary particle consist of half-integer spin that does not undergo strong interactions. Leptons exist on two main classes i.e. charged leptons, and neutral leptons. Electron, electron neutrino, muon, muon neutrino, tau and tau neutrino are the six types of leptons.

8 0

3 years ago

An object is propelled along a straight-line path by a force. If the new force were doubled, is acceleration would

Aneli [31]

Still go straight but would obviously go up in speed!!

Hope this helps plz mark as brainlist and 5 star

6 0

3 years ago

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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).