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Dmitry_Shevchenko [17]

2 years ago

14

The acceleration due to gravity at the north pole of Neptune is approximately 10.7m/s2. Neptune has mass 1026kg and radius 25000

km and rotates once around its axis in a time of about 16h.
Part A

What is the gravitational force on an object of mass 5.5kg at the north pole of Neptune?

Part B

What is the apparent weight of this same object at the Neptune's equator? (Note that Neptune

1 answer:

Scilla [17]2 years ago

7 0

Answer:

(A) Force on the object will be 58.85 N

(B) Apparent weight will be equal to 57.21 N

Explanation:

We have given mass of Neptune m = 1026 kg

Radius R = 25000 Km

Time period = 16 hour

We know that 1 hour = 3600 sec

So 16 hour = 16×86400 = 57600 sec

Acceleration due to gravity at north pole of Neptune $g=10.7m/sec^2$

(A) Mass of the object m = 5.5 kg

So gravitational force on the object $F_g=mg=5.5\times 10.7=58.85N$

(B) Velocity will be equal to $v=\frac{2\pi R}{T}=\frac{2\times 3.14\times25000\times 1000}{57600}=2725.694m/sec$

Apparent weight of the object will be equal to $w=f_g-\frac{mv^2}{r}$

$W=58.85-\frac{5.5\times (2725.694)^2}{25000\times 1000}=57.21N$

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The acceleration is given as:

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For g = 10 m/s^2, we get

a = 10 sin(30°) = 10 * 1/2 = 5 m/s^2

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Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

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Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

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The initial velocity of the wooden block, U = 0 m/s

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Therefore,

mV + MV = mu + MU

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2 years ago

A physics student with a stopwatch drops a rock into a very deep well, and measures the time between when he drops the rocks and

tankabanditka [31]

Answer:

h= 161.06 m

Explanation:

Given that

Speed of the sound ,C= 343 m/s

Total time ,t= 6.2 s

lets take the depth of the well = h

The time taken by stone before striking the water = t₁

we know that

$h=\dfrac{1}{2}gt_1^2$

$t_1=\sqrt{\dfrac{2h}{g}}$

The time taken by sound =t₂

$t_2=\dfrac{h}{343}$

The total time

t = t₁+ t₂

$6.2 = \sqrt{\dfrac{2h}{g}}+\dfrac{h}{343}$

$6.2 = \sqrt{\dfrac{2h}{9.81}}+\dfrac{h}{343}$

Now by solving the above equation we get

h= 161.06 m

Therefore the depth of the well will be 161.06 m.

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2 years ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

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Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

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If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

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