What are the factors that affect the resistance of a wire?

natita [175]

The answer would be c

8 0

2 years ago

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

#LearnwithBrainly

7 0

3 years ago

A large, 34.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of m

Lorico [155]

Answer:

length L of the clapper rod for the bell to ring silently = 0.756m

Explanation:

We are given;

Mass of Bell;m_b = 34 kg

Distance of centre of mass from pivot;d = 0.7m

The bells moment of inertia about an axis at the pivot;I = 18 kg.m²

Mass of clapper;m_c = 1.8 kg

Length of slender rod is L

Now, the formula for period of physical pendulum having small amplitude is given as;

T_b = 2π√(I/mgd)

Where;

I is moment of inertia

m is mass

g is acceleration due to gravity = 9.8 m/s²

d is distance from rotation axis to centre of gravity

Plugging in the relevant values and using mass of bell, we have;

T_b = 2π√(18/(34*9.81*0.7)

T_b = 1.745 s

Now, the formula for period for a simple pendulum which is essentially what the clapper rod is would be;

T_c = 2π√(L/g)

Now, we want to find length of clapper L.

Thus, let's make it the subject;

L = g(T_c/2π)²

Now, we are told that for the bell to ring silently, T_b = T_c.

Thus, T_c = 1.745 s.

So,

L = 9.8(1.745/2π)²

L = 0.756m

7 0

2 years ago

Find the resultant of the following forces: 3.0N east, 4.0N west, and 5.0N in a direction north 60° west.

zzz [600]

Answer:

5.5 N at 50.8° north of west.

Explanation:

To find the resultant of these forces, we have to resolve each force along the x- and y-direction, then find the components of the resultant force, and then calculate the resultant force.

The three forces are:

$F_1=3.0 N$ (east)

$F_2=4.0 N$ (west)

$F_3=5.0 N$ (at 60° north of west)

Taking east as positive x-direction and north as positive y-direction, the components of the forces along the 2 directions are:

$F_{1x}=3.0 N\\F_{1y}=0$

$F_{2x}=-4.0 N\\F_{2y}=0$

$F_{3x}=-(5.0)(cos 60^{\circ})=-2.5 N\\F_{3y}=(5.0)(sin 60^{\circ})=4.3 N$

Threfore, the components of the resultant force are:

$F_x=F_{1x}+F_{2x}+F_{3x}=3.0+(-4.0)+(-2.5)=-3.5 N\\F_y=F_{1y}+F_{2y}+F_{3y}=0+0+4.3=4.3 N$

Therefore, the magnitude of the resultant force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(-3.5)^2+(4.3)^2}=5.5 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{|F_x|})=tan^{-1}(\frac{4.3}{3.5})=50.8^{\circ}$

And since the x-component is negative, it means that this angle is measured as north of west.

7 0

3 years ago

Which three quantities can be used to calculate acceleration?

ikadub [295]

The correct option will be
D. Time, initial velocity and final velocity
The Formula can be written as,
Acceleration=Final velocity-Initial Velocity/Time

5 0

2 years ago