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Sunny_sXe [5.5K]
3 years ago
14

If a 4.5 kg object is dropped from a height of 6.0 m, what will be its velocity when it is halfway toward the ground? (Use g = 9

.80 m/s2, and ignore air resistance.) 7.7 m/s 11 m/s 16 m/s 29 m/s
Physics
2 answers:
dangina [55]3 years ago
7 0
When an object falls or is dropped from rest it's initial velocity is zero.
Using the equations for a motion in straight line. I can find the time it takes to reach 3.0 m down (half way).
x = vt - 4.9t²
-3 = 0 - 4.9t²
-3/-4.9 = t²
0.6122 = t²
0.7825 sec = t

v = v - gt
v = 0 - 9.8(0.7825)
v = -7.67 m/s
the negative denotes downward direction.

You  could also solve the problem using potential and kinetic energy.

Since it starts with maximum PE and gets converted to KE when it hits the ground. mgh = mv²/2
mass cancels, use 3 meters for the halfway distance
-9.8(-3) = v²/2
29.4 * 2 = v²
√(58.8) = 7.67 m/s downwards
notsponge [240]3 years ago
6 0

Answer:

7.7 rounded

Explanation:

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Charge of the alpha particle

Explanation:

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3 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
3 years ago
Infrared light of wavelength 2.5 µm illuminates a 0.20-mm-diameter hole. What is the angle of the first dark fringe in radians?
stepan [7]

Answer:

Θ=0.01525 rad

or

Θ=0.87°

Explanation:

Given data

wavelength λ=2.5 µm =2.5×10⁻⁶m

Diameter d=0.20 mm =0.20×10⁻³m

To find

Angle Θ in radians and degree

Solution

Circular apertures have first dark fringe at

Θ=(1.22λ)/d

Substitute the given values

So

Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m

Θ=0.01525 rad

or

Θ=0.87°

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