All categories

3 years ago

What are the factors that effect the strength of a frictional force???

Physics

1 answer:

seropon [69]3 years ago

5 0

If you write down the formula for friction, you will get an answer.

Ff = u * N Where N is a push down force that an object experiences.

u (mu) is a constant and has no units

It may not be accelerating and still experience friction. A is not correct.

Color and Density will not affect the frictional force. B is not so.

Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.

The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.

You might be interested in

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

4 years ago

Name three variables that affect a life system

sweet-ann [11.9K]

Three things that effect a life system would be
- sunlight
-water
-energy/food

Hope this helps

7 0

3 years ago

Read 2 more answers

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

$\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m$

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}$ (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

$d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m$

hence, the displacement of the airplane is 3.45*10^4 m

6 0

3 years ago

Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S <br>

Tems11 [23]

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

$v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}$

6 0

3 years ago

How does 'g' vary from place to place?

r-ruslan [8.4K]

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

8 0

3 years ago