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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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To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
Determine the magnitude of the electrostatic force on a 0.06000 C charged object when it is placed in an electric field of magni
joja [24]

Answer:

Explanation:

Use the following equation:

E=\frac{q}{F} and solve for F:

F=\frac{q}{E} and filling in:

F=\frac{.0600}{1500}

F = 4.0 × 10^{-4 N

3 0
2 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
answer questions1) The south pole of the steel magnet will be brought closer to the iron nail head. Which pole will be at the po
Anvisha [2.4K]
Answer:

North pole

Explanation:

According to the law of magnetism:

<em>Unlike poles attract while like poles repel</em>

Since the south pole of the steel is brought near the nail, and the nail is meant to attract the steel magnet, the nail domain realigns itself to produce a pole opposite to the pole of the steel magnet brought near it.

Since the North pole is the opposite of the south pole, the North pole will be at the pointed end of the nail so that it can attract the steel magnet.

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Explanation:

1)5.8m/s

2)5.15m/s^2

3)12.69m/s

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8 0
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