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3 years ago

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A point particle of mass m1 = 2.00 kg is at the origin and a second point particle of mass m2 = 6.00 kg is on the x axis at x =

8.0 m. Find the gravitational field at the following locations.a. x = 2.0m b. x = 12.0 m c. Find the point on the x axis for which g = 0.

1 answer:

r-ruslan [8.4K]3 years ago

8 0

Answer:

Ok, the gravitational field in the x-axis can be written as:

g(x) = G*∑mₙ/(xₙ - x)

where the field points in the positive x-axis

where mₙ is the mass of the n-th particle, and xₙ is the position of the n-th particle, then, in our case we have:

g(x) = G*(- 2kg/x + 6kg(8m -x))

then; g(2) = G*( -2kg/2m + 6kg/(8m- 2m)) = G*( -1kg/m + 1kg/m) = 0

g(12) = G*( -(2/12)kg/m +6/(8 - 12)kg/m) = G* (-2/12 kg/m - 6/4kg/m) = -G*(20/12)kg/m

and we already find that the point where g(x) = 0 is 2

this is the x such:

G*(- 2kg/x + 6kg(8m -x)) = 0

then, the thing inside the parentheses must be zero, now we have:

2kg/x = 6kg(8m -x)

then:

x/2kg = (8/6) m/kg - x/(6kg)

x (1/(2kg) + 1/(6kg)) = (8/6)m/kg

x*(4/(6kg)) = (8/6)m/kg

x = (8/6)*(6/4)m = 2m

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A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while

dem82 [27]

12.8 rad

Explanation:

The angular displacement $\theta$ through which the wheel turned can be determined from the equation below:

$\omega^2 = \omega_0^2 + 2\alpha\theta$ (1)

where

$\omega_0 = 0$

$\omega = 34.7\:\text{rad/s}$

$\alpha = 47.0\:\text{rad/s}^2$

Using these values, we can solve for $\theta$ from Eqn(1) as follows:

$2\alpha\theta = \omega^2 - \omega_0^2$

or

$\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}$

$\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}$

$\:\:\:\:= 12.8\:\text{rad}$

7 0

2 years ago

4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav

Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

= 4.5 x 10⁹ Km

= 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

<em>V = d / t m/s</em>

∴ t = d / V

= 4.5 x 10¹¹ m / 3 x 10⁸ m/s

= 15,000 s

= 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

4 0

3 years ago

A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning

Lesechka [4]

Answer:

speed of current is 5 mile/hr

Explanation:

GIVEN DATA:

speed of motorboat = 15 miles/hr relative with water

let c is speed of current

15-c is speed of boat at upstream

15+c is speed of boat at downstream

we know that

travel time=distance/speed

$\frac{10}{15-c} +\frac{10}{15+c} = 1.5$

150+10c+150-10c=1.5(15-c)(15+c)

300=1.5(225-c^2)

300=337.5-1.5c^2

200=225-c^2

c^2=25

c = 5

so speed of current is 5 mile/hr

6 0

3 years ago

I just need x isolated

mamaluj [8]

Answer:

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

Explanation:

$rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0$

Solve using the quadratic formula.

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

7 0

3 years ago

Read 2 more answers

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

e)

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

3 years ago