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During the flight of the ball, what is the direction of its Acceleration due to gravity

vekshin1

Regardless of what direction an object is moving, the acceleration
due to gravity is always directed toward the center of the Earth.
That's the direction commonly known as "down".

4 0

3 years ago

Why do noble gasses rarely react with other elements?

ollegr [7]

Noble gasses have an outer shell full of electrons. A full outer energy level is the most stable arrangement of electrons. As a result, noble gases cannot become more stable by reacting with other elements and gaining or losing valence electrons. Therefore, noble gases are rarely involved in chemical reactions and almost never form compounds with other elements.

3 0

2 years ago

Kinetic energy depends on which two things

Korolek [52]

Answer:

Explanation: relationship between the object and the observer's frame of reference.

5 0

3 years ago

A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0

otez555 [7]

A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
$x(t)=v_0 \cos \alpha t$
$y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
where
- the horizontal motion is a uniform motion, with constant speed $v_0 \cos \alpha$ , where $v_0 = 8.00 m/s$ and $\alpha=20.0^{\circ}$
- the vertical motion is an uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ , initial position h (the height of the building) and initial vertical velocity $v_0 \sin \alpha$ (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
$x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m$

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
$0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
which becomes
$h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m$

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
$y(t)=h-10$
If we substitute this into the equation of y(t), we have
$h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0$
$4.9 t^2 +2.74 t-10 =0$
whose solution is $t=1.18 s$ (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0

3 years ago

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 4.5 m long, its mass is 0.4

lions [1.4K]

Answer:

density = 0.0933 kg/m

speed = 27.581 m/s

Explanation:

given data

length L = 4.5 m

mass m = 0.42 kg

force F = 71 N

to find out

mass density and speed

solution

we find linear mass density

linear mass density = mass / length

put here all value

density = 0.42 / 4.5

density = 0.0933 kg/m

and

speed of wave

speed = √(F/density)

speed = √(0.42/0.933)

speed = 27.581 m/s

7 0

3 years ago