We are given:
m1 = 300 grams
m2 = 500 grams
h1 = h2 = 2 meters
We have to determine the potential energy of both toy cars:
PE = mgh
PE1 = m1gh1
PE1 = 300 grams/1000g/kg * 9.81 m/s^2 * 2 meters
PE1 = 5.886 Joules
PE2 = m2gh2
PE2 = 500 grams/1000g/kg * 9.81 m/s^2 * 2 meters
PE2 = 9.81 Joules
PE2 - PE1 / PE2
9.81 - 5.886 Joules / 9.81 joules * 100 = 40.0%
Therefore, the 500-gram car is greater by 40% in terms of potential energy than the 300-gram car. This is because mass is directly proportional to the potential energy of an object.
Answer:
The product of the decay its Sulfur-32
Explanation:
Phosphorus-32 ( lets write it 
, where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:
where 
its the electron, and 
the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).
So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:
.
.
The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its
.
and the product of such decay its Sulfur-32
Net force = mass x acceleration = 250 x 3 = 750 N
Answer:
d
Explanation:
Correct answer: D). All of the above The particulate control device is a device that is used for the separation of the material that is of a bigger size as compared to filter. In the factory smokestack, the particulate control device filters the particulate matter that is known to cause air pollution. In the vacuum cleaner, a fine dust particle is filtered out. In the laundry dryer, the water and dirt molecule is allowed to pass through it. Hence, all the options are correct.so therefore your answer would be all
Answer:
a. 
b. 
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c. 
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,
- speed of lagging car,
- distance between the cars,
- deceleration of the leading car after braking,
a.
Time taken by the car to stop:
where:
, final velocity after braking
time taken
b.
using the eq. of motion for the given condition:
where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:
is the time taken to stop after braking
