HELP NOWWW Studies based on which of the following were used to describe the current model of the atom?
2 answers:
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Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
13.0m/s
1.2m/s
Explanation:
Given parameters:
Initial speed of the body = 7.1m/s
time taken = 2.23s
Acceleration = 2.64m/s²
Unknown:
Final speed = ?
Solution:
Acceleration is the rate of change of velocity with time.
a =
a = acceleration
V = final speed
U = initial speed
T = time taken
Input the variables and solve for V;
2.64 =
V - 7.1 = 5.9 expression 1
V = 5.9 + 7.1 = 13.0m/s
B
Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;
from expression 1;
V - 7.1 = -5.9
V = -5.9 + 7.1 = 1.2m/s
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Answer:
1.
2.
Explanation:
1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.
In Newton's 2nd law: F = ma
where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.
Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time
Now that acceleration is halved:
You would need to push for twice amount of time
2. The distance traveled by the puck is as the following equation:
So if the acceleration is halved while maintaining the same d:
As , then
. Also
So t increased by 1.14