Answer:
Both objects travel the same distance.
(c) is correct option
Explanation:
Given that,
Mass of first object = 4.0 kg
Speed of first object = 2.0 m/s
Mass of second object = 1.0 kg
Speed of second object = 4.0 m/s
We need to calculate the stopping distance
For first particle
Using equation of motion
Where, v = final velocity
u = initial velocity
s = distance
Put the value in the equation
....(I)
Using newton law
Now, put the value of a in equation (I)
Now, For second object
Using equation of motion
Put the value in the equation
....(I)
Using newton law
Now, put the value of a in equation (I)
Hence, Both objects travel the same distance.
<span>k = 1.7 x 10^5 kg/s^2
Player mass = 69 kg
Hooke's law states
F = kX
where
F = Force
k = spring constant
X = deflection
So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration.
F = kX
F/X = k
115 kg* 9.8 m/s^2 / 0.65 cm
= 115 kg* 9.8 m/s^2 / 0.0065 m
= 1127 kg*m/s^2 / 0.0065 m
= 173384.6154 kg/s^2
Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2
Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So
X/0.39 cm = 115 kg/0.65 cm
X = 44.85 kg/0.65
X = 69 kg
The player masses 69 kg.</span>
Answer:
When extra energy is added
Explanation:
When the ball is released from rest and swings back towards your face, it will only pass closer to the end of the nose as per the initial conditions. However, when extra energy is added to the ball, it strikes the nose since its velocity and heights are increased. Therefore, the only condition under which the ball hits your nose is when extra energy is added to the system.
A. reactants
B. subscript
C. coefficient
D. products
