The options of the question are missing. I have attached it.
Answer:
Satellite in option B will have the greatest speed.
Explanation:
From kepplers third law, we know that
V² = GM/R
Thus, v = √(GM/R)
Where;
v is velocity
G is gravitational constant
M is mass
R is radius
Looking at the options, let's start from the first one;
Option A
Here, mass = (1/2)m and radius = R
So, v = √(GM/R) thus v = √(G(m/2)/R) = √(Gm/2R)
Option B
Here, mass = m and radius = (1/2)R
Thus v = √(Gm/(R/2)) = √(2Gm/R)
Option C
Here, mass = m and radius = R
Thus v = √(Gm/R)
Option D
Here, mass = m and radius = 2R
Thus v = √(Gm/(2R))
Now, inspecting all the options, it's clear that option B will have the greatest velocity because it's numerator is the biggest and will in turn lead to higher velocity.
Answer:
16 J
Explanation:
It is given that,
Work done, W = 2 J
A spring is stretched by 2.0 cm from its equilibrium length
We need to find how much more work will be required to stretch it an additional 4.0 cm.
Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :
The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.
So, the required work done is 16 J.
Well, I guess you can come close, but you can't tell exactly.
It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.
That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.
98.5 m/s = √ [ (horizontal component)² + (vertical component)² ].
The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:
Height = (1/2) (acceleration) (time²) .
If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.
Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
Height = (1/2) (9.81) (10.04)²
= (4.905 m/s²) x (100.8 sec²) = 494.43 meters.
As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.
If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component. That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.
Answer:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.
Explanation:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.
Answer:
d = (75 i ^ + 93 j ^ + 27 k ^) m
, d2 = (900 i ^ + 1116 j ^ + 324 k ^) m
Explanation:
The two objects are in circular orbit together, therefore with the same angular velocity, after the launch they move with the relative velocity, so we can use the kinematic relation
v = d / t
d = v t
Reduce time to units SI
t = 5 min (60 s / 1 min) = 300 s
X axis
x = vₓ t
x = 0.25 300
x = 75 m
Y axis
y = 
t
y = 0.31 300
y = 93 m
Z axis
z= 
t
z = 0.09 300
z = 27 m
d = (75 i ^ + 93 j ^ + 27 k ^) m
For the time of 1 h
t2 = 1 h (3600s / 1 h) = 3600
x2 = 900 m
y2 = 1116 m
z2 = 324 m
d2 = (900 i ^ + 1116 j ^ + 324 k ^) m
