As per Newton's II law we know that

here we know that

so here we will have

so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.
So here correct answer will be
<em>A) force on the object.</em>
Answer: 1433.3 m/min
Explanation:
For 86 Km/h converted to a (m/min), convert kilometers to meters, and hour to minutes
So, 86 Km/h means 86 kilometers per 1 hour
- If 1 kilometer = 1000 metres
86 kilometers = 86 x 1000 = 86,000m
- If 1 hour = 60 minute
1 hour = 60 minutes
In m/min: (86,000m / 60 minute)
= 1433.3 m/min
Thus, 86 Km/h convert to 1433.3 m/min
100 J
Please mark me brainliest it would be greatly appreciated haha
Answer:
D. infinitely extended in all directions
Explanation:
A semi infinite solid is infinitely extended in every direction. It has a single surface and can extend when heat is applied.
The body of a semi infinite solid is idealised, that is, when there is heat present, it expands in all directions to infinity. It can be used for a thick wall because its shape can be changed when subjected to different levels of heat near its surface.
It is also expands as heat is applied because its thickness is negligible.
This idealized body is used for earth, thick wall, steel piece of any shaped quenched rapidly etc indetermining variation of temperature near its surface & other surface being too far to have any impact on the region in short period of time since heat doesn’t have sufficient time to penetrate deep into body thus thickness can be neglected
Answer:
Answered
Explanation:
A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.
W= FS cosθ
θ= 90 ⇒cos90 = 0 ⇒W= 0
B) work done by tension
W= Tcosθ×S= 5cos30×2.30= 10J
C) Work done by friction force
W= f×s=1×2.30= 2.30 J
D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.
E) The net work done= Work done by tension in the rope - frictional work
=10-2.30= 7.7 J