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3 years ago

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Which type of force is the weakest?

1 answer:

Marysya12 [62]3 years ago

8 0

Van der Waals firces

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On level ground a shell is fired with an initial velocity of 34.0 m/s at 51.0° above the horizontal and feels no appreciable air

monitta

Answer:

A. $v_{x}=21.4m/s$

$v_{y}=26.42m/s$

B. $t=2.7s$

C. $y_{max} =35.61m$

D. $x=115.34m$

E. $a_{x}=0$

$a_{y}=-9.8m/s^2=g$

F. $v_{y}=0$

$v_{x}=21.4m/s$

Explanation:

From the exercise our initial values are:

$v_{o}=34m/s$

$\alpha=51º$

A. The horizontal and vertical components are:

$v_{x}=34cos(51)=21.4m/s$

$v_{y}=34sin(51)=26.42m/s$

B. At maximum height the y-component of velocity becomes 0

$v_{y}=v_{o}+a_{y}t$

$0=26.42-9.8m/s^2*t$

$t=2.7s$

C. The maximum height above the ground is:

$v_{y} ^{2}=v_{o}^2+2a_{y}(y-y_{o})$

At maximum height the y-component of velocity becomes 0

$0=(26.42)^2-2(9.8)y$

$y=\frac{-(26.42)^2}{-2(9.8)}=35.61m$

D. To find how dar from its firing point does the sell land we need to calculate how much time does it take to do it first

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

When the shell land y=0

$0=0+(26.42)t-\frac{1}{2}(9.8)t^2$

Solving the quadratic equation for t

$t=0$ or $t=5.39s$

Since time can not be 0 t=5.39s

$x=v_{ox}t=(21.4m/s)(5.39s)=115.34m$

E. Since the velocity at the horizontal component is constant

$a_{x}=0$

The vertical acceleration of the shell is gravity

$g=-9.8m/s^2$

F. At highest point the vertical component is 0. The shell stops going up ans start to go down

$v_{y}=0$

$v_{x}=v_{oy}+a_{x}t$

Since $a_{x}=0$

$v_{x}=21.4m/s$

3 0

3 years ago