____ is a chemical messenger released by virus infected cells

If the coefficient of static friction is 0.40, and the same ladder makes a 51° angle with respect to the horizontal, how far alo

inysia [295]

To solve this problem, we apply the concepts related to the sum of forces and balance in a diagram that will be attached, in order to identify the behavior, direction and sense of the forces. The objective is to find an expression that is in terms of the mass, the angle, the coefficient of friction and the length that allows us to identify when the ladder begins to slip. For equilibrium of the ladder we have,

$\sum F_x = 0$

$\sum F_y = 0$

$\sum M_o = 0$

Now we have that

$f_1 = N_2$

$N_1 = mg$

And for equilibrium of the two forces we have finally

$mgdcos\theta = N_2lsin\theta$

Rearranging to find the distance,

$d = \frac{N_2}{mg}ltan\theta$

$d = \frac{f_1}{mg}ltan\theta$

So if we have that the frictional force is equivalent to

$f_1 = \mu N_1$

$f_1 = \mu mg$

$f_1 = (0.4)(57*9.8)$

$f_1 = 223.44N$

With this value we have that

$d = \frac{(0.4)(57)(9.8)}{57*9.8}(7.5) tan(60\°)$

$d = 5.19m$

Therefore can go around to 5.19m before the ladder begins to slip.

5 0

3 years ago

a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?

Vinil7 [7]

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !

The formula is D = 1/2 a T²

   Distance = (1/2 acceleration) x (time in seconds)²

   Acceleration = (change in speed) / (time for the change)
    = (18 m/s) / (12 sec)
    = 1.5 m/s² .

Distance = (1/2 x 1.5 m/s²) x (12 sec)²
      =     (0.75 m/s²) x (144 sec²) = 108 meters .

5 0

3 years ago

An ice skater weighs 500 [N]. He is coasting to the right at a constant velocity of 2 [m/s]. Assume

luda_lava [24]

Answer:

The net force on the skater is zero. ( $F_{net} = 0\,N$ )

Explanation:

According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. ( $F_{net} = 0\,N$ )

4 0

3 years ago

Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

Inessa [10]

The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$ wires at distance ' r ' is given by $f=\frac{u_0I_1I_2}{2\pi r}$ ....(1) .

If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$