1
2 4. 5 4 5
+3 0 7. 3 0 0
——————
3 3 1 8 4 5
line up the decimal points and add.
hope this helps!
soy de texas, united states
Answer:
<em>The bullet was 0.52 seconds in the air.</em>
Explanation:
<u>Horizontal Motion
</u>
It occurs when an object is thrown horizontally with a speed v from a height h.
The object describes a curved path ruled exclusively by gravity until it hits the ground.
To calculate the time the object takes to hit the ground, we use the following equation:
Note it doesn't depend on the initial velocity but on the height.
The bullet is fired horizontally at h=1.3 m, thus:
t = 0.52 s
The bullet was 0.52 seconds in the air.
Answer:
4
Explanation:
We are given that
K.E at x=0 m=20 J
K.E at x=3 m=11 J
We have to find the value of c.
By work energy theorem
Work done=Change in kinetic energy
W=
![W=[\frac{cx^2}{2}-x^3]^{3}_{0}](https://tex.z-dn.net/?f=W%3D%5B%5Cfrac%7Bcx%5E2%7D%7B2%7D-x%5E3%5D%5E%7B3%7D_%7B0%7D)
The force applied to the second ball by the first ball is 6.734 × 10^-4 N.
<h3>What is impulse of force?</h3>
The impulse of force is defined as the sum of the average force and the duration it is applied.
If the mass of the item remains constant, the impulse of force equals the change in momentum of the object.
Given that: mass of a metal sphere: m = 0.026 kg.
Initial speed of the sphere: u = 3.7 m/s.
When the sphere stops completely, its change in momentum = mu - 0
= 0.026×3.7 N-s.
= 0.0962 N-s.
As the spheres are in contact for 0.007s before the second sphere is shot off down the track, the force applied to the second ball =
change in momentum of 1st ball × time of contact
= 0.0962 × 0.007 N
= 0.0006734 N
= 6.734 × 10^-4 N.
Hence, the force applied to the second ball is 6.734 × 10^-4 N.
Learn more about impulse force here:
brainly.com/question/29787329
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