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

An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth

yanalaym [24]

Note: The answer choices are :

a) Increased

b) Decreased

c) stayed the same

Answer:

The correct option is Increased

The magnitude of the electric field potential difference between the wingtips increases.

Explanation:

The magnitude of the electric potential difference is the induced emf and is given by the equation:

$emf = l (v \times B)$

where l = length

v = velocity

B = magnetic field

As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. $v \times B = vB sin90 = vB\\$ ,

the induced emf = vlB, and thus increases.

The magnitude of the electric field potential difference between the wingtips increases

3 0

3 years ago

nolan ryan has the record for having the speediest fastball in baseball. he could pitch one at 148 i/sec. what is that speed in

Klio2033 [76]

The speed of the ball is 101miles/hr.

A mile is a unit of length that is exactly 1,609.344 metres long. Similarly, 5,280 feet or 1,760 yards make up one mile. The mile is an imperial and common US measurement of distance.

We just have to deal with unit conversions.

One mile is 5280 feet, or 1 ft = 0.000189

The speed of the ball in miles per hour is

$\frac{148ft}{1sec} . \frac{1mile}{5280ft} .\frac{60s}{1min} .\frac{60min}{1hr}$

So, the speed of the ball in miles per hour is 101miles/hr.

Learn more about miles here;

brainly.com/question/23245414

#SPJ4

5 0

1 year ago

A family is skating at an ice rink. The 58.2 kg mother is holding the

MariettaO [177]

Answer:

When I got this question I had to draw it out so if you have to do that, draw 3 stick figures holding hands, one representing the mother, father, and daughter. Then you write their weights on top of them and then draw an arrow pointing from the father to the mother.

Explanation:

use this formula :

$a_{y}$ = $\frac{Fdadshandy}{msys}$

then you fill it in :

$a_{y}$ = $\frac{100N}{35.5kg+58.2kg}$

$a_{y}$ = $\frac{100N}{93.7kg}$

$a_{y}$ = $1.0672$ $m/s^{2}$

then you multiply that with the daughters weight :

$T_{2} x= m_{2} a_{y}$

$T_{2} x = 35.5kg (1.0672 m/s^{2})$

$T_{2} x = 37.89N$

and that's the answer :) : 37.89N

5 0

3 years ago

What best describes the direction of the electric field on a spherical equipotential surface?

Alla [95]

Answer:

Perpendicular to the surface

Explanation:

- Electric field lines represent the direction of the electric field. The electric field lines also correspond to the direction along which the gradient of the electric potential is maximum.

- Equipotentials are lines or surfaces along which the electric potential is constant: the electric potential does not change moving along an equipotential surface.

Given the two definitions, equipotential lines are always perpendicular to the electric field lines. Therefore, in this problem, the direction of the electric field is perpendicular to the spherical equipotential surface.

4 0

3 years ago

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

Tasya [4]

There are 3 forces acting on the stoplight:

• its weight W, with magnitude W = 100 N, pointing directly downward

• two tension forces T₁ and T₂ with equal magnitude T₁ = T₂ = T = 1000 N, both making an angle of θ with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ F = T₁ sin(θ) + T₂ sin(180° - θ) - W = 0

We have sin(180° - θ) = sin(θ) for all θ, so the above reduces to

2T sin(θ) = W

2 (1000 N) sin(θ) = 100 N

sin(θ) = 0.05

θ ≈ 2.87°

If y is the vertical distance between the stoplight and the ground, then

tan(θ) = (15 m - y) / (100 m)

Solve for y :

tan(2.87°) = (15 m - y) / (100 m)

y = 15 m - (100 m) tan(2.87°)

y ≈ 9.99 m

3 0

3 years ago