Explanation:
see the above attachment to solve the question and get the answer.
hope this helps you.
Answer: please see attached work.
Explanation: please see attached work. Assuming 500 sheets of paper = 20 lb. (typicical value).
Answer:
<h3>14.97m/s</h3>
Explanation:
Given
Initial velocity of the car u = 8m/s
Distance travelled by the rider S = 40m
Acceleration a = 2m/s²
Required
rider's velocity after the acceleration v
Using the equation of motion
v² = u²+2as
v² = 8²+2(2)(40)
v² = 64+160
v² = 224
v = √224
v = 14.97m/s
Hence the rider's velocity after the acceleration is 14.97m/s
A) reactants interact to form products with different chemical and physical properties
The image distance can be determined using the mirror equation: 1/f = 1/d_o + 1/d_i, where, f is the focal length, d_o is the object distance, and d_i is the image distance. Given that f = 28.2 and d_o = 33.2 cm, the value of d_i is calculated to be 187.248 cm. On the other hand, the image height is obtained using the magnification equation wherein, h_i/h_o = -d_i/d_o, where h_i is the image height and h_o is the object height. Using the given values, h_i is equal to -26.79 cm. Note that the negative sign indicates that the image is inverted.
