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Delvig [45]
3 years ago
8

Which example is correctly matched with its type of friction?

Physics
1 answer:
vlabodo [156]3 years ago
3 0

Answer:

A. pushing a car that isn't moving

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g A uniform ladder whose length is 4.4 m and whose weight is 330 N leans against a frictionless vertical wall. The coefficient o
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Answer:

Distance = 4.4 [m]

Explanation:

This problem can be easily solved using a static analysis of forces acting on the ladder, taking into account the respective distances. For easy understanding, a free body diagram should be made.

We perform a sum of force on the X-axis equal to zero, to find that the force exerted by the wall is equal to the friction force on the floor.

Then we perform a summation of forces on the Y axis, to determine that the normal force exerted by the floor is equal to the weight of the ladder.

We know that the friction force is equal to the product of normal force by the coefficient of friction.

In this way, by relating the friction force to the equations deduced above we can find the force exerted by the wall.

Then we make a summation of moments around the base point of the ladder, the equation realized can be seen in the attached image.

In the last analysis we can find the relationship between the horizontal and vertical distance of the ladder, with respect to the wall and the floor.

Then with the complementary analysis of the Pythagorean theorem we can find another additional equation.

The result of the greater distance is 4.4 [m]

3 0
3 years ago
A string of 26 identical Christmas tree lights are connected in series to a 120 V source. The string dissipates 73 W. What is th
spin [16.1K]

To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

R_{eq}= \frac{V}{I}

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

P=VI \rightarrow I = \frac{P}{V}

I =0.608 A

Applying Ohm's law

R_{eq} = \frac{120V}{0.608A}

R_{eq} = 197.4\Omega

Therefore the equivalent resistance of the light string is 197.4\Omega

6 0
3 years ago
Now Abel and Kato use what they learned to answer the following problem. The initial speed of a tennis ball is 54 m/s and the la
34kurt

Answer:

h=19.4m

R=199.07 m

Explanation:

To solve this problem we use the parabolic motion equations:

We define:

v_{i}total initial speed  =54\frac{m}{s}

\alpha=angle that forms the total initial speed with the horizontal line= 21°

v_{ix}:initial speed component in horizontal direction

=v_{i} cos\alpha =54*cos 21= 50.41 m/s

v_{iy}:initial speed component in vertical direction

=v_{i} sin\alpha=54*sin21=19.35 m/s

v_{x} :horizontal speed at any point on the parabolic path

v_{y}: vertical speed at any point on the parabolic path

g= acceleration of gravity= 9,8\frac{m}{s^{2} }  

Equation of the speed of the football in the vertical direction :

(v_{y} )^{2} =(v_{yi} )^{2} -2*g*y  Equation (1)

Calculation of the maximum height(h)

The speed of the ball (vy) in the vertical direction gradually decreases until its value is zero when it reaches the maximum height.

We replace y=h, v_{y} =0,  v_{iy} = 19.39\frac{m}{s}, g=9.8\frac{m}{s^{2} } in the equation(1)

0=19.35^{2} -2*9.8*h

h=\frac{19.35^{2} }{2*9.8}

h=19.1 m

Calculating of the range (R)

Fórmula: R=v_{ix} *t   (m)   Equation (2)

R is the maximum horizontal distance the ball reaches.

The time (t) for the ball to reach R is twice the time the ball spends to reach the maximum height. Then, we calculate the time(t_{h} )) when the ball reaches the maximum height

We apply the following equation to calculate t_{h}:

v_{y} =v_{iy} -g*t_{h}

0=19.35-9.8*t_{h}

9.8*t_{h} =19.35

t_{h} =\frac{19.35}{9.8}

t_{h} =1.97 s

t=2*t_{h}

t=3.95 s  

We replace v_{ix} =50.41 \frac{m}{s} and t=3.95\frac{m}{s} in the equation (2)

R=50.41*3.95

R=199.07 m

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3 years ago
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