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dmitriy555 [2]
3 years ago
5

Help me

Physics
1 answer:
mezya [45]3 years ago
4 0
Choose answer C.....
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How long it takes to reach maximum height in seconds
olganol [36]
The dotted path is the path of the ball. it reaches it's maximum height at the top where vertical, y-velocity = 0
The initial y-velocity = 19sin(70°)
initial y-velocity = 17.85 m/s

Use one of the kinematic equations with velocity and time. No displacement because we don't want to worry about figuring that out.

v = u - gt
0 = 17.85 - 9.8t
-17.85 = -9.8t
17.85/9.8 = t
1.82 sec = t




6 0
3 years ago
What derived unit is used to measure the slope of the line in this graph?
vlabodo [156]

Answer:

C. g/cm³

Explanation:

The slope is measured by calculating the variation of the Y values over the X values between two points on a line.  

So, the formula is: Slope = Δy/Δx

That means that we also take the units.

In this case, the Y-axis unit is in g, while the X-axis unit is in cm³.

Dividing a Y-variation over an X-variation will give you g/cm³.

In this case, let's assume the line passes through (10,100) (not exactly, but close enough for the example), and it passes through (0,0)

So the slope would be: (100-0) g / (10-0) cm³ = 10 g/cm³

3 0
3 years ago
What happens to the acceleration of an object if the net force on it increases
Virty [35]
IS any one going to answer this I need the answer also
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6 0
3 years ago
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Which chemical equation is balanced?
scoundrel [369]
The 4na + 02 —> wna20
6 0
2 years ago
A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ s
irina1246 [14]

Answer:

v_{f}\approx 2.097\,\frac{m}{s}

Explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

m_{cyl}\cdot g\cdot y_{o} = \frac{1}{2}\cdot m_{cyl} \left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

g\cdot y_{o} = \frac{1}{2}\cdot\left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

The translational speed is:

v_{f} = \sqrt{\frac{2\cdot g\cdot y_{o}}{\left(1 + \frac{1}{R}  \right)}}

v_{f} = \sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (4.80\,m)}{\left(1 + \frac{1}{0.049\,m}  \right)} }

v_{f}\approx 2.097\,\frac{m}{s}

7 0
3 years ago
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