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2 years ago

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On a drive from one city to? another, victor averaged 39 mph. if he had been able to average 72 ?mph, he would have reached his

destination 11 hrs earlier. what is the driving distance between one city and the? other?

1 answer:

PilotLPTM [1.2K]2 years ago

5 0

Let the unknown distance be xmiles
x/39-x/72=11hr
72x-39x/2808=11hr
33x/2808=11
33x= 30888
x=936miles
U can substitue back to check
at speed of 72mph, he would need 936/72=13hrs
at speed of 39mph, he would need 936/39=24hr
the difference is 24-13=11

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A father racing his son has 1/2 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.4 m

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Answer: a) 0.78 m/s b) 1.57 m/s

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 V^2/v^2 = 1/4

 V = v/2  Second equation:

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The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.

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The resistance is 27.5 ohms

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Two identical blocks, A and B, are on a horizontal surface, as shown above. There is negligible friction between the surface and

e-lub [12.9K]

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the speed of the center of mass stays the same

Explanation:

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A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.50 x 105 P

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Answer:

$F \approx 19.5 N$

Explanation:

From the question we are told that:

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Bottle cap area $A_b= 4.40 * 10-4 m^2$

Generally the equation for Resultant pressure $P_r$ is give as is mathematically given by

$P_r=P_{CO_2}-P_a$

Where

$P_a=atmospheric\ pressure = 1.013*10^5 pa$

$P_r=1.50 * 105 Pa-1.013*10^5 pa$

$P_r=0.487*10^5 pa$

Generally the equation for Force exerted by screw F is give as is mathematically given by

$F = P*A\\F = 0.487*10^5*4.00*10^-4\\ F = 19.48 N$

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1 year ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

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2 years ago