Explanation:
The orbiting period of a satellite at a height h from earth' surface is
T=2πr32gR2
where r=R+h.
Then, T=2π(R+h)R(R+hg)−−−−−−−−√
Here, R=6400km,h=1600km=R/4
T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√
Putting the given values,
T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h
Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.
Let it be nR.Then
T=2π(R+nR)R(R+nRg)−−−−−−−−−−√
=2π(Rg)−−−−−√(1+n)32
=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32
(5075s)(1+n)32=(1.41h)(1+n)32
For T=24h, we have (24h)=(1.41h)(1+n)32
or (1+n)32=241.41=17
or 1+n(17)23=6.61
or n=5.61
The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.
A man applies a force of 540 N to the barrow in a direction 75 from the horizontal. He moves the barrow 30 m along the level ground. Calculate the work he does against friction?
The work done by the man against friction is 4,192.86 J.
The given parameters;
force applied, F = 540 Nangle of inclination, θ = 75⁰horizontal distance, x = 30 mThe work done by the man against friction is calculated as follows;
[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]
Thus, the work done by the man against friction is 4,192.86 J.
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A 2457 kg car moves with initial speed of 18 ms-l. It is stopped in 62 m by its brakes.
What is the force applied by the brakes?
Answer:
Explanation:
The work of the brakes will equal the initial kinetic energy of the car
Fd = ½mv²
F = mv²/2d
F = 2457(18²) / (2(62))
F = 6,419.903...
F = 6.4 kN
which of the following statements might be used to defend the Act of 1848
CAN SOMEONE PLEASE HELP ME
Answer:
she will eventually slow down and come to a stop
d what is
7 A rocket of mass 10000 kg uses 5.0kg of fuel and oxygen
to produce exhaust gases ejected at 5000 m/s. Calculate the
increase in its velocity
Answer:
Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.
Explanation:
Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.
Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.
The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.
Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].
The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].
The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:
[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].
The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:
[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:
[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:
[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].
[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.
[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].
[tex]9995\, \Delta v = 25000[/tex].
[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].
Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.
How would you best define the word drug?
A: Something that makes you tired
B: Something that can kill you
C: Something that effects your body and mind
D: Stored for energy
someone help
Answer:
C
Explanation:
Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body
A gold doubloon 6.1 cm in diameter and 2.0mm thick is dropped over the side of a Pirate Ship. When it comes to rest on the ocean floor at a depth of 770m how much has its volume changed
The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.
The volume of the gold doubloon can be determined by;
volume = [tex]\pi r^{2}[/tex] + h
where r is the radius of the coin and h is its thickness.
Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm
r = [tex]\frac{diameter}{2}[/tex]
= [tex]\frac{61}{2}[/tex]
r = 30.5 mm
Thus,
volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2
= 5847.2857
Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].
Since the gold doubloon is not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.
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