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k0ka [10]
2 years ago
10

A rocket lifts off the pad at cape canaveral. according to newton's law of gravitation, the force of gravity on the rocket is gi

ven by f(x) = − gmm x2 where m is the mass of the earth, m is the mass of the rocket, g is a universal constant, and x is the distance (in miles) between the rocket and the center of the earth. take the radius of the earth to be 4000 miles, so that x > 4000 miles. find the work, w1, done against gravity when the rocket rises 2000 miles.
Physics
1 answer:
Alenkinab [10]2 years ago
8 0
The equation is the Law of Universal Gravitation. The gravitational constant G is equal to 6.67×10⁻¹¹ Nm²/kg². The mass of the Earth is <span>5.972 ×10</span>²⁴ kg. Compared to the mass of the Earth, the mass of the rocket is negligible. So, we don't need to know the mass of the rocket. Substituting the values:

F = (6.67×10⁻¹¹ Nm²/kg²)(5.972 ×10²⁴ kg)/(4000 miles*(1.609 km/1mile))²
F = 9616423.08 N

The work is equal to
W = Fd
W = (9616423.08 N)(2000 miles*1.609 km/mile)
W = 9.095×10¹⁰ Joules
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a bicycle pump contains 20cm3 of air at a pressure of 100kpa the air is then pumped in a single stroke through a valve into a ty
riadik2000 [5.3K]
If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂

Now, we know: 
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre) 
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)

We can solve for p₂:
p₂ = (p₁V₁)/V₂
    = (100×120)/100
    = 120kPa

Therefore your answer is: 120kPa
8 0
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
ziro4ka [17]

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:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

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

7 0
2 years ago
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton. par
Alenkasestr [34]
<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
 We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
 The Coulombs Constant k = 9.0 x 10^9
 Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
  Mass of the Electron = 9.11 x 10^-31
  F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
  ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
 v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
 v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
 Since one orbit is the distance,
  one orbit = circumference = 2 x pi x r; distance s = v x t.
 v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
  t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
 Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
6 0
2 years ago
How can controlling the way light bends and reflects be used to help people?
Vinil7 [7]

too much sun is dangerous for humans and can cause cancer so it's important that light is reflected for example a pool reflects water back to space that is why water sometimes is cold because it reflects light

4 0
2 years ago
A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of
scZoUnD [109]
Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by  50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)

8 0
2 years ago
Read 2 more answers
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