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3 years ago

6

A car travels 70 km in two hours 104 km in the next 1.5 hours and then 79 km in two hours before reaching its destination what w

as the cars average speed?

1 answer:

stiks02 [169]3 years ago

4 0

Average speed = (total distance) / (total time)

Total distance = (70km + 104km + 79km) = 253 km

Total time = (2hr + 1.5hr + 2hr) = 5.5 hrs

Average speed = (253 km) / (5.5 hrs)

<em>Average speed = 46 km/hr</em>

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You are riding a bicycle at 20 m/s at 10:00 am. At 10:01 am, you notice that you are traveling at 40 m/s. You are headed W. What

Juliette [100K]

Answer:

20m/s^2

Explanation:

Acceleration=Change in velocity/time taken for change

40-20/1

20m/s^2

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

3 years ago

A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

Triss [41]

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

Here, Tension= m*g = m*9.81

and linear mass density= $\frac{8.25 g}{65 cm}$

Linear mass density= $\frac{8.25*10^-3}{65*10^-2}$

Linear mass density= 0.0127 kg/m

Velocity= $2*\frac{l}{t}$

Velocity= 2 * $\frac{65*10^-2}{7.84}$

Velocity= 165.8 m/s

So putting all these values in equation we get

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

165.8= $\sqrt{\frac{m*9.81}{0.0127} }$

Solving we get

m= 35.58 kg

or m= 35.6 kg

3 0

3 years ago

A car cruises at a constant rate of 50 miles per hour. How much time will it take to go 600 miles?

Len [333]

Speed is constant. 50 miles = 1 hour. 600/50 = 12. 1hr(12) = 12 hours.

7 0

2 years ago

8. The legs of a young man are each 0.650 meters long. What is his maximum walking speed?

Norma-Jean [14]

Answer:

2.52 m/s

Explanation:

When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.

Sum of the forces in the radial direction:

∑F = ma

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 0.650 m:

v = √(9.8 m/s² × 0.650 m)

v = 2.52 m/s

8 0

3 years ago