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3 years ago

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1. A person is lifting a suitcase into the trunk of a car. Assuming the suitcase is already in the air, draw the FBD and write t

he sum of forces equation for the box. (Assume the person is lifting straight up).

1 answer:

Alik [6]3 years ago

7 0

Answer:

Explanation:Assume you are lifting an object with mass 20 kg from the ground to a height of 1.5 m. Assume that you are exerting a constant force in the upward direction and that you are moving the object upward with uniform velocity. The net force on the object is zero. The force you are exerting is equal in magnitude and opposite in direction to the force of gravity. As you are lifting the object you are doing work on the object.

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A push on a textbook with 10 N moves it to a distance of 2 cm. How much force is need to move it to 4 cm?

uysha [10]

Answer:

20N

Explanation:

Ratio of N to cm-

10:2

so to make 2=4 times 2 so The ratio is now-

20:4

so to move 4 cm you need to push 20N.

8 0

3 years ago

FORCE AND DISPLACEMENT AT AN ANGLE A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes a 25.0° angle

Bingel [31]

Answer: 6117.58 J

Explanation:

We know that W=Fd*cos(theta) where theta is the angle between the displacement and the force.

In this case, we are given that F=225 N, d=30 m, and theta=25 degrees.

Plugging all this in we get

W=225*30*cos(25)=6117.58 J

7 0

3 years ago

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

7 0

4 years ago

The brightest star in the night sky in the northern hemisphere is Sirius. Its distance from Earth is estimated to be 8.7 light y

katen-ka-za [31]

Answer: $5.11(10)^{13}miles$

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.

This unit is equivalent to $5.879(10)^{12}miles$ , which mathematically is expressed as:

$1Ly=5.879(10)^{12}miles$

Doing the conversion:

$8,7Ly.\frac{5.879(10)^{12}miles}{1Ly}=5.11(10)^{13}miles$ This is the distance from Earth to Sirius in miles.

4 0

3 years ago

Which example shows an endothermic reaction?

Nata [24]

Answer:

reactants --> products + thermal energy

Explanation:

Heat is absorbed in endothermic reaction.

6 0

3 years ago