Answer:
Time needed: 2.5 s
Distance covered: 31.3 m
Explanation:
I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by
v2f=v2i−2⋅a⋅d
Isolate d on one side of the equation and solve by plugging your values
d=v2i−v2f2a
d=(15.02−10.02)m2s−22⋅2.0ms−2
d=31.3 m
To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation
vf=vi−a⋅t, which will get you
t=vi−vfa
t=(15.0−10.0)ms2.0ms2=2.5 s
Answer:
-252.52
Explanation:
L = Distance between lenses = 10 cm
D = Near point = 25 cm
= Focal length of objective = 0.9 cm
= Focal length of eyepiece = 1.1 cm
Magnification of a compound microscope is given by
The angular magnification of the compound microscope is -252.52
Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.
Explanation: Please see the attachments below
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
