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slavikrds [6]
3 years ago
15

What happened if there is no frictional force?​

Physics
1 answer:
Komok [63]3 years ago
4 0

Answer: Friction stops things from sliding apart. If there was no friction everything would slide to the lowest point. With no friction the only possible movement would be falling to a lower point under gravity.

Explanation: In a frictionless world, more objects would be sliding about, clothes and shoes would be difficult to keep on and it would be very difficult for people or cars to get moving or change direction. Students should be encouraged to consider how dependant their world is on the beneficial action of friction. Humans and other objects will become weightless without gravity. If we have no gravity force, the atmosphere would disappear into space, the moon would collide with the earth, the earth would stop rotating, we would all feel weightless, the earth would collide with the sun, and as a consequence.

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How does inertia affect someone who isn't wearing a seatbelt
Paul [167]
Basically, when someone is resting in an accelerated vehicle without restraint from a seatbelt, the force of stopping the vehicle will be when inertia occurs, and that force of the vehicle coming to a stop will affect the passenger (without a seatbelt/restraint from another force or object) greatly by throwing them.
For example;
If I were to be riding in a vehicle (without a seatbelt) that's accelerating at 40 m/s^2 and it suddenly gets slammed on the breaks, I will be thrown forward from inside the vehicle.

I hope this helps!
7 0
4 years ago
A box with the mass of 20 kg at 5 m is lifted to 20 m. How much work was 7 points<br> done?
kirza4 [7]

Answer:

2,900\: \mathrm{J}

Explanation:

Work is given by the equation W=F\Delta x where F is force and \Delta x is displacement.

Displacement is defined as change in position. In this case, the box moves from 5m to 20m. Its displacement is 20-5=15\:\mathrm{m}.

The force acting on the box is the force of gravity, given as F_g=mg.

Therefore, the total work done is:

W=F\Delta x=mg\Delta x = 20\cdot 9.81\cdot 15=2,943=\fbox{$2,900\:\mathrm{J}$}(two significant figures).

3 0
3 years ago
A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi
Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put  the value into the formula

-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2

u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}

u=-2.496\ m/s

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

v = u+gt

Put the value in the equation

-2.496=0-9.8\times t

t =\dfrac{2.496}{9.8}

t=0.254\ sec

The total time is

t'=t+1.95

t'=0.254+1.95

t'=2.204\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times9.8\times(2.204)^2

s=23.80\ m

Hence, (a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

4 0
3 years ago
How are earthquakes distributed on the map
professor190 [17]
On the tectonic plates
6 0
3 years ago
If the horizontal component of a vector is 6 m/s and the vertical component is also 6 m/s, what is the resultant value of the ve
BlackZzzverrR [31]
Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.
6 0
4 years ago
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