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Irina-Kira [14]
3 years ago
12

An 1800-kg truck pulls a 710-kg trailer away from a stoplight with an acceleration of 1.20 m/s2 . Part A What is the net force e

xerted by the truck on the trailer? Express your answer to three significant figures and include the appropriate units. Assume that the positive x axis is directed toward the direction of motion.
Physics
1 answer:
faust18 [17]3 years ago
8 0

Answer:

F = 852 N

Explanation:

We apply Newton's second law to the trailer :

F = m*a   Formula (1)

F : net force exerted by the truck on the trailer Newtons (N)

m : mass of the trailer in kilograms (kg)

a : acceleration of the trailer in meters over second square (m/s²)

Data

a=1.20 m/s² : acceleration of the trailer

m=710 kg : mass of the trailer

We replace data in the   Formula (1)  to calculate the net force exerted by the truck on the trailer

F = (710 kg)*(1.20 m/s²)

F = 852 N

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Answer:

C

Explanation:

The weight will always be different while mass is described as the stuff inside an object, and that stays the same.

Such as it weighs differently in space.

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A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w
steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

h = L - L\cos{31°} = L(1 - \cos{31°})

so its initial gravitational potential energy U_0 is

U = mgh = mgL(1 - \cos{31°})

\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})

\:\:\:\:\:=0.23\:\text{J}

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U

We know that the initial kinetic energy K_0, as well as its final gravitational potential energy U are zero so we can write the conservation law as

mgL(1 - \cos{31°}) = \frac{1}{2}mv^2

Note that the mass gets cancelled out and then we solve for the velocity v as

v = \sqrt{2gL(1 - \cos{31°})}

\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}

\:\:\:\:\:= 1.3\:\text{m/s}

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3 years ago
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A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters
emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

F = \frac{mv^{2} }{r}

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = 2\pi r

Work on a particle = 2πrF = 2\pi mv^{2}

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.

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3 years ago
A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h
MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

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the friction force F_s= μmg= 0.3×55×9.81

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to move the ball horizontal force required is 161.86 N

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Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu
lora16 [44]

The initial force between the two charges is given by:

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where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

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So the force has increased by a factor 6.

8 0
3 years ago
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