Answer:
Because there's buoyancy force
Explanation:
What is wrong here is that he assumes that gravity is the only force affecting the weight of the balloon on the scale here. In reality there's also buoyancy force which lifts the full blown balloon upwards.
A better type of container would be an air tank where its volume does not change when it's empty and when it's full of air.
A. The acceleration during the slide is 6.86 m/s²
B. The time taken to slide until he stops is 1.2 s
<h3>How to determine the force of friction</h3>
- Mass (m) = 81.5 Kg
- Coefficient of friction (μ) = 0.7
- Acceleration due to gravity (g) = 9.8 m/s²
- Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
- Frictional force (F) =?
F = μN
F = 0.7 × 798.7
F = 559.09 N
<h3>A. How to determine the acceleration</h3>
- Mass (m) = 81.5 Kg
- Frictional force (F) = 559.09 N
- Acceleration (a) =?
a = F / m
a = 559.09 / 81.5
a = 6.86 m/s²
<h3>B. How to determine the time </h3>
- Initial velocity (u) = 8.23 m/s
- Final velocity (v) = 0 m/s
- Decceleration (a) = -6.86 m/s²
- Time (t) =?
a = (v – u) / t
t = (v – u) / a
t = (0 – 8.23) / -6.86
t = 1.2 s
Learn more about acceleration:
brainly.com/question/491732
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ANSWER:
What effect does a catalyst have on a system in equilibrium?
The system is unaffected.
~batmans wife dun dun dun....
It would be C. the color of the pot. its pretty obvious that i would not effect the project.
Answer:
c. 48 cm/s/s
Explanation:
Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with a mass of 2M when acted upon by a net force of 2F?
from newtons second law of motion ,
which states that change in momentum is directly proportional to the force applied.
we can say that
f=m(v-u)/t
a=acceleration
t=time
v=final velocity
u=initial velocity
since a=(v-u)/t
f=m*a
force applied is F
m =mass of the object involved
a is the acceleration of the object involved
f=m*48.........................1
in the second case ;a mass of 2M when acted upon by a net force of 2F
f=ma
a=2F/2M
substituting equation 1
a=2(M*48)/2M
a=. 48 cm/s/s
