Answer:_COC1\/2+_H\/2O>_HC1+CO\/2
Explanation:
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The horizontal speed of the object 1.0 seconds later is 1) 5.0 m/s.
Explanation:
The motion of an object thrown horizontally off a cliff is a projectile motion, which follows a parabolic path that consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
This means that the horizontal speed of an object in projectile motion does not change, and remains constant during the whole motion.
Since in this case the object has been launched with a horizontal speed of
v = 5.0 m/s
this means that this speed will remain constant during the motion, so its horizontal speed 1.0 s later is also 5.0 m/s.
Learn more about projectile motion:
brainly.com/question/8751410
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Answer: Approximately 8.0g of water
Explanation:
Answer:
a) ![W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J](https://tex.z-dn.net/?f=W_%7Bg%7D%3Dmdx%20%3D%200.21%20kg%20%2A9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%200.10m%3D0.2058%20J)
b) ![W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J](https://tex.z-dn.net/?f=W_%7Bspring%7D%3D%20-%5Cfrac%7B1%7D%7B2%7D%20Kx%5E2%20%3D-%5Cfrac%7B1%7D%7B2%7D%20200%20N%2Fm%20%280.1m%29%5E2%3D-1%20J)
c) ![V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s](https://tex.z-dn.net/?f=V_i%20%3D%5Csqrt%7B2%20%5Cfrac%7BW_g%20%2B%20W_%7Bspring%7D%7D%7B0.21%20kg%7D%7D%7D%3D%5Csqrt%7B2%20%5Cfrac%7B%281-0.2058%29%7D%7B0.21%20kg%7D%7D%7D%3D2.75m%2Fs)
d)
or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:
![W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J](https://tex.z-dn.net/?f=W_%7Bg%7D%3Dmdx%20%3D%200.21%20kg%20%2A9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%200.10m%3D0.2058%20J)
Part b
For this case first we can convert the spring constant to N/m like this:
![2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}](https://tex.z-dn.net/?f=2%20%5Cfrac%7BN%7D%7Bcm%7D%20%5Cfrac%7B100cm%7D%7B1m%7D%3D200%20%5Cfrac%7BN%7D%7Bm%7D)
And the work donde by the spring on this case is given by:
![W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J](https://tex.z-dn.net/?f=W_%7Bspring%7D%3D%20-%5Cfrac%7B1%7D%7B2%7D%20Kx%5E2%20%3D-%5Cfrac%7B1%7D%7B2%7D%20200%20N%2Fm%20%280.1m%29%5E2%3D-1%20J)
Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:
![W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i](https://tex.z-dn.net/?f=%20W_%7Bg%7D%20%2BW_%7Bspring%7D%20%3D%20K_%7Bf%7D%20-K_%7Bi%7D%3D0-%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2_i)
And if we solve for the initial velocity we got:
![V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s](https://tex.z-dn.net/?f=V_i%20%3D%5Csqrt%7B2%20%5Cfrac%7BW_g%20%2B%20W_%7Bspring%7D%7D%7B0.21%20kg%7D%7D%7D%3D%5Csqrt%7B2%20%5Cfrac%7B%281-0.2058%29%7D%7B0.21%20kg%7D%7D%7D%3D2.75m%2Fs)
Part d
Let d1 represent the new maximum distance, in order to find it we know that :
![-1/2mV^2_i = W_g + W_{spring}](https://tex.z-dn.net/?f=-1%2F2mV%5E2_i%20%3D%20W_g%20%2B%20W_%7Bspring%7D)
And replacing we got:
![-1/2mV^2_i =mg d_1 -1/2 k d^2_1](https://tex.z-dn.net/?f=-1%2F2mV%5E2_i%20%3Dmg%20d_1%20-1%2F2%20k%20d%5E2_1)
And we can put the terms like this:
![\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20k%20d%5E2_1%20-mg%20d_1%20-1%2F2%20m%20V%5E2_i%20%3D0)
If we multiply all the equation by 2 we got:
![k d^2_1 -2 mg d_1 -m V^2_i =0](https://tex.z-dn.net/?f=%20k%20d%5E2_1%20-2%20mg%20d_1%20-m%20V%5E2_i%20%3D0)
Now we can replace the values and we got:
![200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0](https://tex.z-dn.net/?f=200N%2Fm%20d%5E2_1%20-0.21kg%289.8m%2Fs%5E2%29d_1%20-0.21%20kg%285.50%20m%2Fs%29%5E2%29%20%3D0)
![200 d^2_1 -2.058 d_1 -6.3525=0](https://tex.z-dn.net/?f=200%20d%5E2_1%20-2.058%20d_1%20-6.3525%3D0)
And solving the quadratic equation we got that the solution for
or 18.3 cm because the negative solution not make sense.
Explanation:
Red, green, and blue are therefore called additive primaries of light. ... When you block two lights, you see a shadow of the third color—for example, block the red and green lights and you get a blue shadow. If you block only one of the lights, you get a shadow whose color is a mixture of the other two.
First, your definition of a shadow is incorrect. A shadow is an area that receives less light than its surroundings because a specific source of light is blocked by whatever is "casting" the shadow. Your example of being outside reveals this. The sky and everything around you in the environment (unless you are surrounded by pitch black buildings) is sending more than enough light into your shadow, to reveal the pen to your eyes. The sky itself diffuses the sunlight everywhere, and the clouds reflect plenty of light when they are not directly in front of the Sun.
If you are indoors and have two light bulbs, you can throw two shadows at the same time, possibly of different darknesses, depending on the brightness of the light bulbs.
It can take a lot of work to get a room pitch black. One little hole or crack in some heavy window curtains can be enough to illuminate the room. There are very few perfectly dark shadows.