Ask question

Ask question

All categories

3 years ago

15

A loading car is at rest on a track forming an angle of 25° with the vertical. The gross weight of the car and its load is 5500

lbf, and it is applied at a point 30 in. from the track, halfway between the two axles. The car is held by a cable attached 24 in. from the track. Determine the tension in the cable and the reaction at each pair of wheels (that is determine the forces exerted by the track surface on the wheels of the car; note that wheels are part of the car, do not consider them separately

1 answer:

koban [17]3 years ago

4 0

Idk I just need other answers

You might be interested in

A box rests on top of a flat bed truck. the box has a mass of m = 18 kg. the coefficient of static friction between the box and

konstantin123 [22]

For this case, the first thing you should do is write the kinematic motion equation of the block.
We have then:
vf = vo + a * t
Where,
vf: Final speed.
vo: Initial speed.
a: acceleration.
t: time.
Substituting the values:
(16) = (0) + a * (16)
Clearing the acceleration:
a = 16/16 = 1m / s ^ 2
Note: the other data for this case are not used in this problem.
answer:
The acceleration of the box is 1m / s ^ 2

5 0

3 years ago

B. A 20kg wagon is pulled up a 30° incline at an acceleration of 2.5ms?. The force pulling the wagon is parallel to the incline

yawa3891 [41]

The value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

<h3>Coefficient of the kinetic friction</h3>

The value of coefficient of kinetic friction is calculated as follows;

F - Ff = ma

F - μmgcosθ = ma

where;

F is applied force
μ is coefficient of kinetic friction
m is mass of the wagon
a is acceleration of the wagon

182 - μ(20 x 9.8 x cos30) = 20(2.5)

182 - 169.74μ = 50

182 - 50 = 169.74μ

132 = 169.74μ

μ = 132/169.74

μ = 0.78

Thus, the value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

Learn more about coefficient of friction here: brainly.com/question/20241845

7 0

2 years ago

Differentiate between concave and convex lens. Write your own answer

Katyanochek1 [597]

Answer:

A <em>concave</em><em> </em><em>lens</em><em> </em><em>is</em><em> </em><em>thinner</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>cen</em><em>ter</em><em> </em><em>and </em><em>thick</em><em>er</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>edges</em><em> </em><em>while</em><em> </em><em>a</em><em> </em><em>convex </em><em>lens </em><em>is</em><em> </em><em>thicker</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>and</em><em> </em><em>thinner</em><em> </em><em>at</em><em> </em><em>the</em><em> edges</em><em>.</em>

6 0

2 years ago

Read 2 more answers

Why are citizens obligated to respond to such documents?

Zina [86]

To protect the constitutional right to confront ones accused

5 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago