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Airida [17]
3 years ago
9

In order to start an object moving, you must first overcome the force of _________ friction.

Physics
2 answers:
navik [9.2K]3 years ago
7 0
It would be static friction which is what you have to overcome when an object is not in motion. When you move an object friction works against it like gravity and air resistance. I hope this helps!
Luda [366]3 years ago
6 0

Answer:

Static friction.

Explanation:

Static friction is defined as the force of friction  that keeps an object at rest position.

This friction must be overcome to start an object moving . If an object is in motion, it will experiences a force of friction called  kinetic friction.

If a force is applied which is of small magnitude to an object, the static friction will posses or applies an equal magnitude which is in opposite direction of applied force.

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3 years ago
If you push a 200 kg box with a horizontal force of 1,172 N and kinetic friction resists the motion with a force of 962 N, what
Lynna [10]

Answer:

a = 1.05m.s²

Explanation:

Fnet = m×a

Fapplied - friction = m×a

1172 - 962 = 200 × a

210 = 200a

a = 1.05

5 0
1 year ago
The velocity of a passenger relative to a boat is -vpb. The velocity of the boat relative to the river it is moving on is vbr. T
RideAnS [48]

Answer:

vps = vbr + vrs - vpb

Explanation:

  • If the passenger were at rest, his speed relative to the shore will be identical to the boat's, as follows:
  • vps = vbr + vrs
  • As he is moving in a direction opposite to the boat's, his velocity relative to the shore must be less than if he were at rest, in the same quantity that he was moving opposite to the boat, as follows:
  • vps = vbr+ vrs -vpb
5 0
2 years ago
The movement of water is able to transport minerals and nutrients. Which statement best explains why water is able to do this?
Marrrta [24]

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3 0
3 years ago
Read 2 more answers
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
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