The weight of a person increase when the elevator is going up.
<h3>
Weight of the person in the elevator</h3>
The weight of the person in the elevator is calculated as follows;
<h3>When the person is going up</h3>
F = ma + mg
F = m(a + g)
where;
- a is acceleration of the person
- g is acceleration due to gravity
<h3>When the person is going down</h3>
F = mg - ma
F = m(g - a)
Thus, the weight of a person increase when the elevator is going up.
Learn more about weight here: brainly.com/question/2337612
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Answer:
C. At a particular instant
Explanation:
Speed is the defined as the ratio between the distance covered by an object and the time taken:
where d is the distance and t the time.
However, there are two possible measurements of speed:
- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is
- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.
Molecular Geometry is the three-dimensional structure or arrangement of atoms in a molecule.
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Answer:17.08 s
Explanation:
Given
distance between First and second Runner is 45.6 m
speed of first runner=3.1 m/s
speed of second runner=4.65 m/s
Distance between first runner and finish line is 250 m
Second runner need to run a distance of 250+45.6=295.6 m
Time required by second runner
time required by first runner to reach finish line
Thus second runner reach the finish line 80.64-63.56=17.08 s earlier