Answer:
The speed of the comet at the surface of the star is approximately 1,208,694.7 m/s
Explanation:
Question parameter obtained online; The mass of the star, M = 5 × 10³¹ kg
Explanation;
The given mass of the comet, m = 2 × 10⁸ kg
The initial velocity of the comet, v → 0
The distance of the comet from the star, d = 700,000,000 km
The gravitational potential at d = G·M·m/d
The kinetic energy of the comet, K.E. = m·v²/2
The kinetic energy of the comet at d = m·(0)²/2 = 0
The gravitational potential at the surface of the star, R = G·M·m/R
The kinetic energy of the comet at the surface of the star, R = m·(v)²/2 = 0
Where;
M = The mass of the star = 5 × 10³¹ kg
[tex]M_{Sun}[/tex] = The mass of the Sun = 1.989 × 10³⁰ kg
M/[tex]M_{Sun}[/tex] = 5 × 10³¹/(1.989 × 10³⁰) ≈ 25
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
R = The radius of the star
Therefore, we have;
m·(0)²/2 - G·M·m/d = m·v²/2 - G·M·m/R
∴ v = √((G·M·m/R - G·M·m/d)×2/m) = √(2·G·M(1/R - 1/d))
Therefore; v = (2 × 6.67430 × 10⁻¹¹ × 5 × 10³¹ × (1/R - 1/700,000,000,000))
v = 81696389149.1×√(1/R - 1/700,000,000,000).
The speed of the comet at the surface of the star, v = 81696389149.1×√(1/R - 1/700,000,000,000)
The mass radius relationship is given as follows;
[tex]\dfrac{R}{R_{Sun}} = 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]
[tex]R = R_{Sun} \times 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]
The radius of the Sun = 696,340,000 M
∴ R ≈ 696,340,000 × 1.3 × √(25.14) = 4538865694.76
R = 4538865694.76 m
v = 81696389149.1×√(1/4538865694.76 - 1/700,000,000,000) ≈ 1208694.7 m/s
An oscillator completes 240 cycles in 5.2 minutes.
Calculate its period (in seconds) and frequency (in Hz).
Answer:
I. Period = 1.3 seconds
II. Frequency = 0.769 Hertz
Explanation:
Given the following data;
Number of oscillation = 240 cycles
Time = 5.2 minutes.
Conversion:
1 minute = 60 seconds
5.2 minutes = X seconds
X = 60 * 5.2
X = 312 seconds
To find the following;
I. Period
Mathematically, the number of oscillation of a pendulum is given by the formula;
[tex] Number \; of \; oscillation = \frac {Time}{Period} [/tex]
Making period the subject of formula, we have;
[tex] Period = \frac {Time}{Number \; of \; oscillation} [/tex]
Substituting into the formula, we have;
[tex] Period = \frac {312}{240} [/tex]
Period = 1.3 seconds
II. Frequency
[tex] Frequency = \frac {1}{Period} [/tex]
Substituting the values into the formula, we have;
[tex] Frequency = \frac {1}{1.3} [/tex]
Frequency = 0.769 Hertz
Angie walked a distance of 90 meters east in 70 seconds. What was her
velocity?
A. 0.78 m/s east
B. 1.3 m/s east
O C. 7 m/s east
D. 9 m/s east
1) The position of an object to the north of a flagpole is given by x(t) = bt2 – c , where b and c are constants.
a) What is v(t), the velocity of the object as a function of time?
b) What is a(t), the acceleration of the object as a function of time?
c) At some time t the object is located at the flagpole. What is the velocity of the
object at that instant?
Answer:
a) The velocity of the object as a function of time, v(t) is 2·b·t
b) The acceleration of the function of time, a(t) is 2·b
c) The time at which the object is at the flagpole is t = √(c/b)
Explanation:
The function that gives the position of the object north of the flagpole, x(t) is presented as follows;
x(t) = b·t² - c (b and c are constants)
a) The velocity of the object as a function of time, v(t), is derived as follows
v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t
The velocity of the object as a function of time, v(t) = 2·b·t
b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b
c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;
At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c
∴ 0 = b·t² - c, which gives
b·t² = c
t² = c/b
t = ±√(c/b), we reject the negative value to get;
The time at which the object is at the flagpole, t = √(c/b).
If 20N force produces an acceleration of 5ms^-2 In a body then the mass of the body will be:
A.4kg
B.5kg
C.1/4kg
D.1/5kg
A cyclist goes round a circular path of circumference 343 m in s. The angle made by him, with the vertical is
WITH STEPS PLZ
you need the number of seconds to calculate the distance. I'm assuming that the seconds were written in the question but you forgot to write the here, you can calculate the distance then and find the exact point where the cyclist stopped since you have the circumference. then you can find the angle (after finding the distance he cycled) by subtracting it from 360 since the circular path will have an angle of 360 degrees.
Although your question lacks some data A general answer is provided :
The angle made vertically = 360° - x°
where x = angle at which the the cyclist stopped
and circumference of the circular path = Total Distance travelled by the cyclist
First step : determine the distance travelled by the cyclist vertically
circumference = 2*π*r
343 = 2*π * r
∴ r = 343 / ( 2π ) = 54.59
therefore distance travelled by the cyclist vertically = 54.59 * 2 = 109.18 m
Given that the time travelled is missing
assuming the angle to the distance travelled by the cyclist before it stopped vertically = 360° - x°
This is because the Total angle of a circular path = 360°
Learn more about circular paths : https://brainly.com/question/24210487
determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3
Answer:
divide the density of solution by density of water
EXPLANATION:
LIKE:
1.25÷1000kgm-3
A piece of gum becomes stuck upon a skateboard's wheel. What is the centripetal acceleration of the piece of gum if the wheel's radius is 30 mm and the tangential velocity is 0.5 m/s?
Answers:
a. 8.33 m/s^2
b. 0.5 m/s^2
c. 30 mm/s^2
d. 0.33 m/s^2
e. 83.20 m/ft^2
Answer:
a
Explanation:
a_c = v_t^2/r
a_c = (0.5)^2/0.03
a_c = 8.33 m/s^2
Can someone do this for the football
Estimated volume
(cm3)
Estimated density
(g/cm3)
edge.
Answer:
0.10 g/cm3
TRUST ME GUYS
the masses of your hand and your notebook are quite small, so the force of attraction between them is
Two football players run towards each other along a straight path in Penrith Park in the clash between the Melbourne storms and the Penrith Panthers a month ago. Melbourne's Justin Olam who is about 95kg and ran towards Viliame Kikau at 3.75m/s. Viliame Kikau is 111kg and moves towards Justin Olam at 4.10m/s. They end up in a head-on collision and are stuck together.
A) What is their velocity immediately after the collision?
B) What are the initial and final kinetic energies of the system?
Answer:
a) v = 0.4799 m / s, b) K₀ = 1600.92 J, K_f = 5.46 J
Explanation:
a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.
Initial instant. Before the crash
p₀ = m v₁ + M v₂
where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive
Final moment. After the crash
p_f = (m + M) v
as the system is isolated, the moment is preserved
p₀ = p_f
m v₁ + M v₂ = (m + M) v
v =[tex]\frac{m v_1 + M v_2}{m+M}[/tex]
let's calculate
v = [tex]\frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}[/tex]
v = 0.4799 m / s
b) let's find the initial kinetic energy of the system
K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2
K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2
K₀ = 1600.92 J
the final kinetic energy
K_f = ½ (m + M) v ^ 2
k_f = ½ (95 + 111) 0.4799 ^ 2
K_f = 5.46 J
7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.
Answer:
8392
Explanation:
d=s/t
A force of 5N accelerates a mass by 2 m/s². What will be the acceleration if the force and mass were increased to twice their original value?
Answer:
4m/s4
Explanation:
4. When setting goals, you should do everything EXCEPT which of the following?
Take into account your current level of activity.
Ask all of your friends what they think your goals should be..
Set a time limit.
Be realistic.
An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
Answer:
35
Explanation:
dfddffffffffffffdddfr
Có một số điện trở giống nhau R0 = 3
. Cần ít nhất bao nhiêu điện trở để có một
đoạn mạch có điện trở Rtđ = 8
Answer:
hlo
Explanation:
hlo olz mark me as brainlest
Question in the picture please help me...
Show all steps please....
Answer: i)A to B : (ice) freezing
ii) B to C (water) boiling
C to D (steam) evaporating
explanation: 0° is the freezing point of water when temperature increases from 0° the water starts melting. As 100° is the boiling point of water so at 100° the water completely melts and it starts boiling during boiling water changes into steam(water vapour) and it evaporates
write down any 5 example of conservation of momentum?
Answer:
1) Motion of air mass moving from equator northward (closer to earth axis)
2) Motion of object in orbit
3) Collision of 2 objects
4) Skater changing rotation by extension of arms
5) Motion of rocket due to velocity of expelled gas
Một vật không mang điện sẽ bị nhiễm điện dương khí
Answer:
không có điện
Explanation:
Can someone pls help, thank you in advance!
What is an example of a force applied at an angle to displacement
Answer:
an object sliding down hill
Explanation:
On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.
An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)
transition metals
noble gases
metalloids
alkali metals
Answer:
Alkali metals
Explanation:
Elements in this group are highly reactive, soft, lustrous and highly conductive.
An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.
Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.
This causes them to be soft and highly reactive because:
The single valance electron leads to weak bonds amongst the element's atoms which makes them softThe elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.Examples of alkali electrons include:
Lithium Sodium Potassium etcIn conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.
Find out more at https://brainly.com/question/18722874.
State the relative position for the earth and sun in a lunar eclipse (in a partial and total eclipse)
Answer:
A lunar eclipse is when the Earth passes between the moon and the sun, casting a shadow on the moon. This can only occur when the sun, Earth and moon are aligned exactly, or very closely so, with the Earth in the middle.
Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iron
Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up
what is Newton's first law of motion?
EXPLAIN WITH SOME EXAMPLES
Answer:
The tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia.
Example.: A ball at rest on the ground continues to be at rest unless someone kicks it or any external force acts on it.
Explanation:
I hope this will help you buddy
una caja en reposo se traslada 93 cm con un peso de 67N en un tiempo de 9,89h.¿cual es la aceleración la masa y la fuerza de dicho objeto
Answer:
a. Acceleration, a = 1.47 * 10^{-9} m/s²
b. Mass = 4.57 * 10^{10} kilograms
c. Force = 67.12 Newton
Explanation:
Given the following data;
Distance = 93 cm to meters = 93/100 = 0.93 meters
Weight = 67 N
Time = 9.89 hours to seconds = 35604 seconds
Initial velocity = 0 m/s (since it's starting from rest)
Acceleration due to gravity, g = 9.8 m/s
a. To find the acceleration, we would use the second equation of motion;
[tex] S = ut + \frac{1}{2} at^{2} [/tex]
Where;
S is the distance covered or displacement of an object.
u is the initial velocity.
a is the acceleration.
t is the time.
Substituting the values into the equation, we have;
[tex] 0.93 = 0*35604 + \frac{1}{2} * a*35604^{2} [/tex]
[tex] 0.93 = 0 + \frac{1}{2} * 1267644816a [/tex]
[tex] 0.93 = 633822408a [/tex]
[tex] Acceleration, a = \frac{0.93}{633822408} [/tex]
Acceleration, a = 1.47 * 10^{-9} m/s²
b. To find the mass
Weight = mass * acceleration due to gravity
67 = mass * 1.47 * 10^{-9}
[tex] Acceleration, a = \frac{67}{1.47 * 10^{-9}} [/tex]
Mass = 4.57 * 10^{10} kilograms
c. To find the force;
Force = mass * acceleration
Force = 4.57 * 10^{10} * 1.47 * 10^{-9}
Force = 67.12 Newton
difference between effort distance and load distance
Answer:
Lever systems are simple machines that change or increase the input force that we apply to the load. The lever provides us with some mechanical...
Answer:
● Effort arm or Effort distance (ED): The perpendicular distance from the fulcrum to the point of effort is called effort arm.
● Load arm or Load distance (LD): The perpendicular distance from the fulcrum to the point of load is called load arm.
Động vật nào sau đây máu đi nuôi cơ thể không pha trộn giữa máu giàu O2 và máu giàu CO2?
A.
Bò sát, chim, thú
B.
Cá, bò sát, chim
C.
Cá, lưỡng cư, bò sát
D.
Cá, chim, thú
Answer:
D
Explanation:
Cá xương, chim, thú, cá sấu không có sự pha trộn máu giàu O2 và máu giàu CO2 ở tim vì tim cá có 2 ngăn, tim các loài chim, thú, cá sấu có 4 ngăn
please answer
during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled the last 120 km in 3.2 hours. calculate the average speed of the car during the journey.
average speed of the car is 23.9 km/h
What is your wheel and axle
Explanation:
The wheel and axle is a type of simple machine used to make tasks easier in terms of manipulating force by applying the concept of mechanical advantage.
A lady walks 10 m to the north, then she turns and continues walking 30 m due east.
Determine her(a) distance covered
(b) displacement.
Answer:
The distance covered is 40 m and the displacement is 31,6m.
Explanation:
The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .
Which element would have the lowest electronegativity? (1 point)
an element with a small number of valence electrons and a large atomic radius
an element with a small number of valence electrons and a small atomic radius
an element with a large number of valence electrons and a small atomic radius
an element with a large number of valence electrons and a large atomic radius
Answer:
an element with a small number of valence electrons and a large atomic radius
