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3 years ago

Someone help me! you get 27 points if you get it right

Physics

1 answer:

pashok25 [27]3 years ago

8 0

Answer:

1. D

2. D

3. A

The reason why your body goes right and the car car goes left is because your body tries to stay where it was, which is on the right. Your upper body doesn't feel a force and because of that continues in the same direction. Your lower half is pulled out from under you by seat friction to the left, leaving you leaning to the right.

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A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

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