Answer:
Yes, it will float.
Explanation:
It is given that,
A liquid has a density of 0.85 g/cm³.
We know that, the density of water is 1 g/cm³.
The density of the liquid is less than that of water. If the density of the liquid is less than that of water, it will float in it. Hence, if you pour some of the liquid into a glass of water, it will float.
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
![v=15 m/s](https://tex.z-dn.net/?f=v%3D15%20m%2Fs)
- radius of the hill:
![r=100 m](https://tex.z-dn.net/?f=r%3D100%20m)
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,
![m \frac{v^2}{r}](https://tex.z-dn.net/?f=m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
, so we can write:
![mg-N=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg-N%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
(1)
By rearranging the equation and substituting the numbers, we find N:
![N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%28975%20kg%29%289.81%20m%2Fs%5E2%29-%28975%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D7371%20N%20%20)
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
![N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%2862%20kg%29%289.81%20m%2Fs%5E2%29-%2862%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D469%20N%20)
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
![mg=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
from which we find
Cause it’s denser at the bottom
The question involves the knowledge of kinematics and dynamics. The answers are;
a) Time taken to stop the car = 10 s
b) The operation that slows the car is Friction
c) The size of the force = 1200 N
<h3>
What is Deceleration ?</h3>
Deceleration is the opposite of acceleration. When an object is going to rest, it will be decelerating and the final velocity will be equal to zero.
Given that the driver of a car moving at 15 m/s along a straight level road applies the brakes. The car decelerates at a steady rate of 3/2 m/s²
a) How long does the car take to stop can be found by using the formula
v = u - at
Where
Substitute all the parameters into the formula
0 = 15 - 1.5t
1.5t = 15
t = 15/1.5
t = 10 s
b) The description of the operation that slows the car down after the brake pedal is pressed is simply friction.
c) If the mass of the car is 800. The size of the force slowing the car down will be F = ma
F = 800 × 1.5
F = 1200 N
Therefore, the time taken to stop the car is 10 s and the force slowing the car down is 1200 N
Learn more about Acceleration here: brainly.com/question/605631
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10pi is the answer. i believe. i hope this helps.