What are three elements that are in the same group

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]

Answer:

(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

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

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

c) The work done by the gravitational force on the crate

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

d) the work done by the pull on the crate from the rope

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

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

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

3 years ago

HELP ME ASAP!!!! PIC IS DOWN BELOW!

vagabundo [1.1K]

I think it is B Because it sounds like the best choice

7 0

2 years ago

Read 2 more answers

PLS HELP

Ahat [919]

"This was the idea that non-living objects can give rise to living organisms."

7 0

3 years ago

A 100 mH inductor whose windings have a resistance of 6.0 Ω is connected across a 12 Vbattery having an internal resistance of 3

Alexus [3.1K]

Answer:

The store energy in the inductor is 0.088 J

Explanation:

Given that,

Inductor = 100 mH

Resistance = 6.0 Ω

Voltage = 12 V

Internal resistance = 3.0 Ω

We need to calculate the current

Using ohm's law

$V = IR$

$I=\dfrac{V}{R+r}$

Put the value into the formula

$I=\dfrac{12}{6.0+3.0}$

$I=1.33\ A$

We need to calculate the store energy in the inductor

$U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times100\times10^{-3}\times(1.33)^2$

$U=0.088\ J$

Hence, The store energy in the inductor is 0.088 J

7 0

2 years ago

A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppos

brilliants [131]

Answer:

v₂=- 34 .85 m/s

v₁=0.14 m/s

Explanation:

Given that

m₁=70 kg ,u₁=0 m/s

m₂=0.15 kg ,u₂=35 m/s

Given that collision is elastic .We know that for elastic collision

Lets take their final speed is v₁ and v₂

From momentum conservation

m₁u₁+m₂u₂=m₁v₁+m₂v₂

70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂

70 x v₁ + 0.15 x v₂=5.25 --------1

v₂-v₁=u₁-u₂ ( e= 1)

v₂-v₁ = -35 --------2

By solving above equations

v₂=- 34 .85 m/s

v₁=0.14 m/s

4 0

3 years ago