The radius of the satellite's motion is 2.49 × 10⁷ m
Since the centripetal force, F = mv²/r equals the gravitational force, F' = GMm/r² on the satellite,
F = F'
mv²/r = GMm/r²
v² = GM/r
Making r subject of the formula, we have
r = GM/v² where r = radius of satellite's motion, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of earth = 5.972 × 10²⁴ kg and v = speed of satellite = 4,000 m/s.
Substituting the values of the variables into the equation, we have
r = GM/v²
r = 6.67 × 10⁻¹¹ Nm²/kg² × 5.972 × 10²⁴ kg/(4,000 m/s)²
r = 39.83324 × 10¹³ Nm²/kg ÷ 16 × 10⁶ m²/s²
r = 2.4895775 × 10⁷ m
r ≅ 2.49 × 10⁷ m
So, the radius of the satellite's motion is 2.49 × 10⁷ m.
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Answer:
25.0 million meters
Explanation:
what makes us think that the star system cygnus x-1 contains a black hole?
Answer:
Explanation:
What makes us think that the star system Cygnus X-1 contains a black hole? It emits X-rays characteristic of an accretion disk, but the unseen star in the system is too massive to be a neutron star.
help me solve please
Answer:
please check the image
Explanation:
I hope this image helps you please follow the steps. Thank you
Answer:
1kg/```````2
Explanation:
solution
1 point
What is the acceleration of a bicyclist moving at a constant speed of 10
m/s for 5 seconds? *
PLEASE HELP HURRY
Answer:
0
Explanation:
The bicyclist is not speeding up or slowing down so there is no acceleration.
Calculate the force applied if 100 N/m² pressure is exerted over the area of 0.2m²
Answer: 100 pascals times 0.2 m^2= 20 N force is applied.
Explanation:
a
(2) A 800 g block is pushed up an inclined plane (angled at 18°) with a velocity of 11.8 m/s. The first block slides up the
incline a distance of 2.2 m and strikes a second block with a mass of 300 g also moving at 3.4 m/s up the incline.
The two blocks hit and stick together. Determine the following:
(i) The maximum vertical height of the two blocks when they stop.
(ii) The time needed for the blocks to reach the bottom of the incline after the moment of impact.
(u = 0.19)
this is my attachment answer hope it's helpful to you
Anna used a rock to drive a peg into the ground to put up her tent. If the rock applied a 9. 5 N force in 0. 50 s, what is the impulse on the peg? 4. 8 N • s 9. 5 N • s 10 N • s 19 N • s.
Answer:
4.8 N*s
Explanation:
P = F*t
P = Impulse/ momentum
F = force (Newtons)
t = time (seconds)
P = 9.5N*0.5s
P = 4.75 N*s
Hi there!
[tex]\large\boxed{I = 4.8 N}[/tex]
Impulse = Force (N) × Time (s)
We can also express this as:
I = Δp = Ft
Plug in the given values:
I = 9.5(.50) = 4.75 ≈ 4.8 Ns
A force acts on an object. Which option describes an action that could prevent the object from moving
Answer:
Friction or tension
Explanation:
Friction stops an object from moving in the presence of force
how do astronomers determine the origin of a meteorite that reaches earth?
Researchers begin their study by identifying the rock type and then dating the meteorite
Explanation:
.
solve for the BMI weight 58kg Height 1.61
Answer:
Explanation:
BMI= weight/(height × height) ; weight in kilogram and height in metter
= 58kg / (1.61m × 1.61m )
= (58/ 2.5921) kg/[tex]m^{2}[/tex]
= 22.375 kg/[tex]m^{2}[/tex]
≈ 22.4 kg/[tex]m^{2}[/tex]
What is the correct definition of rarefaction
Answer:
Explanation:
A decrease in the density of something is rarefaction. ... Most of the time, rarefaction refers to air or other gases becoming less dense. When rarefaction occurs, the particles in a gas become more spread out. You may come across this word in the context of sound waves.
a box takes 350 N to start moving the coefficient of static friction is 0.35. what is the weight of the box?
Answer:
101.937 kg
Explanation:
The force needed to get the box moving must just cancel the static friction force:
F = µsmg = 0.35•m•9.81 = 350 ---> m = 350 / (0.35•9.81) = 101.937 kg
Again, with units shown, and using 1 N = 1 kg•m/s2:
0.35•m•9.81(m/s2) = 350 N
solving for m:
m = 350 N / (0.35•9.81 m/s2) = (350 kg•m/s2) / (3.434 m/s2 ) = 101.937 kg
______________________
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Which object has the most thermal energy?
A. A 6 kg rock at 10°C
B. A 10 kg rock at 10°C
C. A6 kg rock at 15°C
D. A 10 kg rock at 15°C
Answer:
D is the answer
Explanation:
D is the most highest one so
the answer is D
Answer:
10 kg rock at 15 degrees celcius
Part IV Objects on an incline w/ Tension + Friction
1. A small 63 kg sleigh is pulled by a rein attached to horse up a 15'angle hill to the
horizontal. The tension of the rein is 510 N. The coefficient of kinetic friction is
0.25
a. What is the normal force that the sleigh exerts on the hill?
b. What are the magnitude and direction of the sleigh's acceleration?
(a) The normal force on the sleigh is 596.36 N.
(b) The magnitude and direction of acceleration of the sleigh is 3.2 m/s² upwards.
The given parameters;
mass of the sleigh, m = 63 kginclination of the hill, θ = 15⁰tension on the rein, F = 510 Ncoefficient of friction, μ = 0.25The normal force on the sleigh is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 63 \times 9.8 \times cos(15)\\\\F_n = 596.36 \ N[/tex]
The magnitude and direction of acceleration of the sleigh is calculated as follows;
[tex]\Sigma F= ma\\\\F - mgsin(\theta) - F_f = ma\\\\F - mgsin(\theta) - \mu F_n = ma\\\\510\ - \ 63 \times 9.8 \times sin15 \ -\ 0.25\times 596.36 = 63a\\\\201 .11 = 63a\\\\a = \frac{201.11}{63} \\\\a = 3.2 \ m/s^2 \ upwards[/tex]
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What mass M will cause this system accelerate e 0.75 m/s²?
Answer:
5.54 [kg].
Explanation:
1) if m₁=6; g=9.81; a=0.75 and m₂ - reqired mass, then
2) it is possible to write: m₁*a=m₁*g-m₂*g, and
3) [tex]m_2=\frac{m_1(g-a)}{g}; \ => \ m_2=\frac{6*(9.81-0.75)}{9.81}=5.54[kg].[/tex]
A long-distance runner runs at a constant speed of 4.8 m/s. How long does it take them to run 1.5 km?
Convert km to meters:
1km = 1000 m
1.5 km x 1000 = 1500 m
1500m / 4.8 m/s = 312.5 seconds
312.5 seconds / 60 seconds per minute =5.2 minutes = 5 minutes 12.5 seconds
Find the velocity of a body of mass 100 g having kinetic energy of 20 J.
Answer:
Explanation:
Let v be the velocity of object ,
Given that mass of object , m=100 g=
1000
100
kg=0.1 kg
& kinetic energy of object =20 J
K.E =
2
1
mv
2
20=
2
1
×0.1v
2
⟹v
2
=400
⟹v=20 m/s
how many tries did it take to invent the lightbulb?
3. If John leaves his house to walk his dog and he walks around the block, what is
his distance?
180 Meters
180 Meters
180 Meters
180 Meters
Answer:
180 meters
Explanation:
a soccer ball is kicked at an angle of 35° and it lands on even ground.
A) What angle will produce the same range?
B) Which angle of the two will produce the highest ball?
C) Which angle of the two will produce the ball that is in the air the longest?
D) What angle in this situation would produce the furthest range?
Hi there!
A)
The angle that will produce the same range is the compliment of 35°.
Thus, kicking the ball at 55° will result in the same range.
We can prove this by using the derived range equation:
[tex]R = \frac{v^2sin2\theta}{g}[/tex]
An angle of 35° yields:
[tex]R = \frac{v^2sin(2*35)}{g} = .939R[/tex]
An angle of 55° yields:
[tex]R = \frac{v^2sin(2*55)}{g} = .939R[/tex]
Both are the same, thus indicating that 55° produces the same range.
B)
The angle of 55° will produce the higher ball because the VERTICAL component of the ball's velocity is greater compared to kicking the ball at 35° degree.
sin(55) > sin(35)
C)
The angle of 55° will result in the ball being in the air the longest because when a ball is in the air (assuming no air resistance), the ball experiences an acceleration due to gravity of -9.8 m/s², causing the vertical velocity to decrease until it eventually reaches 0 m/s at the top of its path. A greater initial vertical velocity means that it will take longer for the ball to fall.
We can prove this using:
vf = vi + at
0 = vy - 9.8t
vy/9.8 = t
Greater vy (vertical component of velocity) ⇒ greater time taken.
D)
The angle that would result in the furthest range is 45°.
We can prove this using calculus. Recall the above range equation:
[tex]R = \frac{v^2sin2\theta}{g}[/tex]
We can take the derivative and use the first-derivative test to find its critical point:
[tex]\frac{dR}{d\theta} = \frac{v^22cos2\theta}{g} = 0[/tex]
Evaluate:
[tex]v^22cos2\theta = 0 \\\\cos2\theta = 0 \\\\2\theta = 90^o\\\\\boxed{\theta = 45^o}[/tex]
A. The angle that will produce the same range as a 35° kick is 53°.
B. The angle that produces the highest ball is 45°.
C. The angle that produces the ball that is in the air the longest is 45°.
D. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.
A. The range of a projectile is the horizontal distance it travels before it hits the ground. The range of a projectile is determined by the initial velocity, the angle of projection, and the acceleration due to gravity.
For a given initial velocity, the range of a projectile is maximized when the angle of projection is 45°. However, if the ground is not level, the range of a projectile can be maximized at other angles.
In the case of a soccer ball kicked at an angle of 35°, the range will be maximized at an angle of 53°. This is because the range of a projectile is maximized when the vertical component of the initial velocity is equal to the horizontal component of the initial velocity.
The angle of 53° is the angle that produces a vertical component of the initial velocity that is equal to the horizontal component of the initial velocity when the ball is kicked at an angle of 35°.
Therefore, the angle that will produce the same range as a 35° kick is 53°.
B. The angle that produces the highest ball is 45°.
C. The angle that produces the ball that is in the air the longest is 45°.
D. The angle that produces the furthest range depends on the initial velocity and the acceleration due to gravity. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.
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Consider the schematic of the molecule shown, with two hydrogen atoms, H, bonded to an oxygen atom, O. The angle between the two bonds is 106°. If the bond length r = 0.103 nm long, locate the center of mass of the molecule. The mass mH of the hydrogen atom is 1.008 u, and the mass mO of the oxygen atom is 15.9999 u. (Use a coordinate system centered in the oxygen atom, with the x-axis to the right and the y-axis upward. Give the coordinates of the center of mass in nm.)
The definition of the center of mass allows to find the result for the position of the mass center of more than the H₂O molecule is;
[tex]x_{cm} = 0 \ and \ y_{cm} = 6.9 10^{-3 } nm[/tex]
the concept of center of mass of a system is the point where external forces are applied, it is given by the expression
[tex]\frac{x}{y} =\frac{1}{M_{total}} \sum m_i r_i[/tex]
Where M is the total mass of the systemr_i and m_i sums the position and masses of the element i of the system
In the attachment we have a diagram of the system where the axis and coordinates of the molecules are shown, in this case it is indicated that the origin is in the oxygen atom, so its distance is zero.
[tex]r_{cm} = \frac{1}{2 m + M} \ (2 m r )[/tex] )
They indicate the mass of the hydrogen atom m = 1.008 amu, the bond length r = 0.103 nm and there is an angle 106º between the two hydrogens, therefore the angle from the vertical is:
θ = 106/2 = 53º
Let's find the position of the center of mass for each axis.
x-axis
[tex]x_{cm} = \frac{1}{2m+ M} \ ( m x_1 - m x_2)[/tex]
y-axis
[tex]y_{cm} = \frac{1}{2m + M} \ ( m y_1 + m_2)[/tex]
Let's use trigonometry to find the components of the bond length.
cos θ = [tex]\frac{y}{L}[/tex]
sin θ = [tex]\frac{x}{L}[/tex]
y = L cos θ
x = L sin θ
We substitute.
[tex]x_{cm} = \frac{1}{2m+M} \ (mL (sin 53 + sin (-53)) \\y_{cm} = \frac{1}{2m + M} \ ( mL cos 53 + mL sin 53)[/tex]
we use.
sin θ = - sin -θ
cos θ = cos -θ
Let's calculate.
[tex]x_{cm} = 0 \\y_{cm} = \frac{1}{2 \ 1.008 + 15.9999} \ ( 2 \ 1.008 \ 0.103 cos 53)[/tex]
We see that the center of mass is on the x axis and at a distance from the y-axis of 6.9 10-3 nm
In conclusion using the definition of the center of mass we can find the result for the position of the center of mass of the H₂O molecule is;
[tex]x_{cm}=0[/tex] and [tex]y_{cm}[/tex]cm = 6.9 10⁻³ nm
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SOMEONE PLEASE HELP ME IM GOING INSANE 4 Select the correct answer. A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled 2.0 x 107 meters and is only 10 meters away from its starting point. What is the numerical value of the satellite's average velocity after that one hour? O A. -3.77 x 10-2 meters/second OB. -2.77 x 10-3 meters/second OC. -2.62 x 10-2 meters/second D. -5.55 x 103 meters/second
Hi there!
Since we are dealing with such large distances, we can approximate.
Recall that the equation for speed is:
displacement / time = velocity
OR:
d/t = v
Begin by converting one hour to seconds:
1 hr = 3600 sec
Now, we can solve for velocity:
(-2.0 × 10⁷) / 3600 = -5555.6 m/s ⇒ D: -5.55 × 10³ m/s
why the max Static frictional force is a Little bit bigger than the sliding frictional force
The weight of a cart with a mass of 150 kg is ___ N. (Use 9.8
m/s2 for the acceleration due to gravity.)
Answer:
1470 N
Explanation:
Weight = m*g
So 150 kg * 9.8 m/s^2 = 1470 N
hannah drove 360 mi in 5.2 hours. What was her average speed?
A teacher has two radioactive sources, A and B.
Source A has a longer half-life than source B.
What can be deduced about the nuclei in source A compared with the nuclei
in source B?
Do not refer to isotopes in your answer.
Answer:
The nuclei of source have greater stability than those of source B.
Explain why we draw straight lines to show rays of light?
Answer:
Because light always travels in straight line.
Explanation:
The fact that light travels can be demonstrated by putting an object in its path. If the object is opaque the result is a degree of blackness on the other side of it which is due to the absence of the light. ... As light does appear to travel in straight lines then light is usually modelled as straight lines in drawings.
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when a single star with a mass equal to the sun dies, it will become a
When a single star with a mass equal to the sun dies, it will become a: white dwarf.
A star is a giant celestial (astronomical) object which comprises a luminous sphere of plasma bounded together by its own gravitational force.
A single star can be defined as a type of star that moves through space alone.
Generally, a star is mainly composed of two (2) hot gases and these are:
I. Hydrogen (H).
II. Helium (He).
Based on Astronomy, when a star runs out of hydrogen, depending on its mass, it would either turn into any of the following:
A planetary nebula.A supernova.A white dwarf.Similarly, when a single star with a mass equal to the sun dies, it will turn into a white dwarf.
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If the resistor is connected in series, the voltage across the wire changes. (true or false)
What happens to the volume of a fixed amount of gas if both the pressure and the absolute temperature are doubled
Answer: no change in volume
Explanation:
Which statement indicates that motion has occurred?
A. The reference point has changed.
B. The position of the object has changed.
C.The object has not changed.
D. The object being described has changed