Answer:
a.
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
b.
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]
c.
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]
d. 1.144 × 10¹³ N
Explanation:
The universal law of gravitation is presented as follows;
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
The given mass of the scientific satellite, m = 1,300 kg
The height of the orbit of the satellite, r = 200 km above the Earth's surface
The length of the radius of the Earth, R = 6378 km
The mass of the Earth = 5.97 × 10²⁴ kg
a. The formula for the universal law of gravitation is presented as follows;
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
Where;
M = The mass of the Earth = 5.97 × 10²⁴ kg
m = The mass of the satellite = 1,300 kg
r = The distance between the Earth and the satellite = R + r = 6,378 km + 200 km = 6,578 km
G = The Gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
b. Plugging in the values from the problem into the formula gives;
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]
c. Solving gives;
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]
The force acting between the Earth and the satellite, F ≈ 1.144 × 10¹³ N
d. 1.144 × 10¹³ N
8. A satellite weighs 200 newtons on the surface of
Earth. What is its weight at a distance of one Earth
radius above the surface of Earth?
1) 800 N
3) 100 N
2) 400 N
4) 50 N
The weight of the satellite at a distance of one Earth radius above the surface of Earth is 40 N.
The given parameters:
Weight of the satellite, F = 200 NOne Earth radius above the surface of Earth = 2RThe gravitational force on the satellite is calculated by applying Newton's law of universal gravitation is calculated as;
[tex]F = \frac{Gm_1m_2}{R^2} \\\\F_1 R_1^2 = F_2R_2^2\\\\F_2 = \frac{F_1 R_1^2 }{R_2^2} \\\\F_2 =\frac{200 \times R_1^2}{(2R_1)^2} \\\\F_2 = \frac{200 \times R_1^2 }{4R_1} \\\\F_2 = 40 \ N[/tex]
Thus, the weight of the satellite at a distance of one Earth radius above the surface of Earth is 40 N.
Learn more about Newton's law of universal gravitation here: https://brainly.com/question/9373839
On which planet would your weight be the most and the least?
a.Jupiter and Mercury
B.Jupiter and Neptune
c.Saturn and Neptune
D.
Saturn and Uranus
Answer: on jupiter you would weigh the most and on mars you would weigh the least
Explanation:
If the net external torque acting on the particle is zero, the it satisfies - 1st condition of equilibrium - 2nd condition of equilibrium - Both 1st & 2nd condition of equilibrium - 2nd law of motion
Answer:
a) the distances are zero, Both 1st & 2nd condition
c) the torques are equal but of the opposite sign, 2nd condition of equilibrium
Explanation:
The equilibrium conditions are
1 translational
∑ F = 0
2 rotational
∑ τ = Σ (F_i x r_i) = 0
They tell us that external torque is zero.
Therefore we have two various possibilities
a) the distances are zero, in this case we have a pure translation movement
for this situation the two equilibrium relations are fulfilled
b) the forces are zero, there is no movement
It does not make sense to use the equilibrium relations since there are no forces
c) the torques are equal but of the opposite sign, the forces are on the opposite side of the body.
In this case the 2 equilibrium relation is fulfilled, but not the first one that the force has the same direction
convert 4 kilograms into grams with process
A car traveling 77 km/h slows down at a constant 0.48 m/s2 just by "letting up on the gas." A) Calculate the distance the car coasts before it stops. B)Calculate the time it takes to stop. C)Calculate the distance it travels during the first second. D)Calculate the distance it travels during the fifth second.
Answer:
(a) 477 m
(b) 44.6 s
(c) 21.16 m
(d) 19.24 m
Explanation:
initial speed, u = 77km/h = 21.4 m/s
acceleration, a = - 0.48 m/s2
final speed v = 0
(a) let the stopping distance is s.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s\\\\0 = 21.4^2 - 2 \times 0.48\times s\\\\s = 477 m[/tex]
(b) Let t is time.
Use first equation of motion
v = u + at
0 = 21.4 - 0.48 t
t = 44.6 s
(c) Let the distance is s in first second.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 21.4\times 1 - 0.5\times 0.48\times 1\\\\s = 21.4 - 0.24 = 21.16 m[/tex]
(d) distance traveled in 5 th second is given by
[tex]s = u + 0.5 a (2 n - 1) \\\\s = 21.4 - 0.5\times 0.48 \times (2\times 5 -1)\\\\s= 21.4 - 2.16 = 19.24 m[/tex]
What is a cyclotron??
What is cyclotron frequency?
Answer:
A cyclotron is a type of particle accelerator.
Explanation:
Cyclotron frequency is the frequency of a charged particle moving perpendicular to the direction of a uniform magnetic field B, since that motion is always circular, the cyclotron frequency is given by equality of centripetal force and magnetic Lorentz force.
state the principle of quantization of charge
Answer:
Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge
Answer: Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge. Thus, an object's charge can be exactly 0 e, or exactly 1 e, −1 e, 2 e, etc., but not, say, 12 e, or −3.8 e
Explanation:
PLEEEASEEEEEE HELPPPPP PLEASEEEEEEEEEE science
Answer:
A
Explanation:
transverse wave
Hope this helps you!
Thermodynamics
Sad
064105 adt
What is energy required to melt (10.0 g) mass of aluminum that
is at its melting point? Knowing that the heat of fusion of
aluminum is (380 kJ/kg.)
Answer:
E = 3.8 kJ
Explanation:
Given that,
The mass of the object, m = 10 g = 0.01 kg
The heat of fusion of aluminum is 380 kJ/kg
We need to find the energy required to melt the mass of the aluminium. It can be calculated as follows:
E = mL
So,
E = 0.01 × 380
E = 3.8 kJ
So, the energy required to melt the mass is equal 3.8 kJ.
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that's according to ma observation
describe 2 ways you could increase the efficiency of a household central heating system
Answer:
Arrange an annual service. Treat your boiler like your car. ...
Keep your boiler clean. ...
Bleed your radiators. ...
Top up the pressure. ...
Use a Powerflush. ...
Insulate your pipes. ...
Turn the heating on. ...
If all else fails…
Explanation:
Define second class lever
Answer:
Please find detailed explanation of second class levers below
Explanation:
Levers are one of the classes of machine that possesses three levels namely: first class, second class and third claas. A second class lever is the level of levers in which the load (L) is in between the pivot (F) and the effort (E).
Examples of second class levers include; wheelbarrow, a bottle opener etc. In the bottle opener for example, the bottle lid (load) is in between the pivot of the opener and the hand opening it (effort).
A 62.0 kg water skier at rest jumps from the dock into a 775 kg boat at rest on the east side of the dock. If the velocity of the skier is 4.50 m/s as she leaves the dock, what is the final velocity of the skier and boat?
2.78 m/s to the east
2.78 m/s to the west
0.360 to the west
0.360 to the east
Answer:
The final velocity of the skier and boat is 0.33 m/s to the east.
Explanation:
We can find the final velocity of the skier by conservation of linear momentum:
[tex] m_{s}v_{s_{i}} + m_{b}v_{b_{i}} = m_{s}v_{s_{f}} + m_{b}v_{b_{f}} [/tex]
Where:
[tex]m_{s}[/tex]: is the mass of the water skier = 62.0 kg
[tex]m_{b}[/tex]: is the mass of the boat = 775 kg
[tex]v_{s_{i}}[/tex]: is the initial velocity of the skier = 4.50 m/s (as she leaves the dock)
[tex]v_{b_{i}}[/tex]: is the initial velocity of the boat = 0 (it is at rest)
[tex]v_{s_{f}}[/tex]: is the final velocity of the skier =?
[tex]v_{b_{f}}[/tex]: is the final velocity of the boat =?
Since the final velocity of the skier is the same that the velocity of the boat ([tex]v_{f}[/tex]) we have:
[tex] m_{s}v_{s_{i}} + 0 = v_{f}(m_{s} + m_{b}) [/tex]
[tex]v_{f} = \frac{m_{s}v_{s_{i}}}{m_{s} + m_{b}} = \frac{62.0 kg*4.50 m/s}{62.0 kg + 775 kg} = 0.33 m/s
Therefore, the final velocity of the skier and boat is 0.33 m/s to the east.
I hope it helps you!
If a serve hits the net, what happens?
• Re-do
• Dead Ball
• Other teams point
• Play the ball
Answer:
Depends: Is it tennis or volleyball?
Explanation:
If it's volleyball, it's C, the other teams point.
If it's tennis, it's A, you get a re-do.
A stone is thrown upward with a kinetic energy of 10J. If it goes up to a maximum height of 5 m, find the initial velocity and mass of the stone
Answer:
Explanation:
Begin by remembering that at the stone's max height, the final velocity there is 0. Being in possession of that information along with the fact that the pull of gravity due to acceleration is -9.8 m/s/s, we can use the equation
[tex]v^2=v_0^2+2a[/tex]Δx where v is the final velocity, v₀ is the initial velocity, a is the pull of gravity, and Δx is the displacement. Filling in:
[tex]0^2=v_0^2+2(-9.8)(5.0)[/tex]
0 = v₀² - 98 so
98 = v₀² and
v = 9.9 m/s Now we can put that into the KE equation.
[tex]10=\frac{1}{2}m(9.8)[/tex] and
[tex]m=\frac{2(10)}{98}[/tex] so
m = .20 kg
when is the acceleration of body is positive negative and zero?
Answer:
【 In vector form , if angle between velocity vector and acceleration vector is less than 90° and greater than 0° then it is positive acceleration and if it is less than 180° and greater than 90° then it is negative acceleration. If there is no acceleration vector then it is called zero acceleration. 】
A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward. the velocity is zero, the acceleration is zero, and the force of gravity acting on the ball is zero. the velocity is zero, the acceleration is zero, and the force of gravity acting on the ball is directed downward. the velocity and acceleration reverse direction, but the force of gravity on the ball remains downward. the velocity, acceleration, and the force of gravity on the ball all reverse direction.
Answer:
the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward
Explanation:
Is this exercise in kinematics
v = v₀ - g t
where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.
At the highest point
velocity must be zero.
The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.
The force of the earth on the ball is towards the center of the Earth, that is, down
all other alternatives are wrong
WILL MARK BRAINLIEST
Answer:
152,155 J
Explanation:
115,333 + 36,822 = 152,155J
Object A has twice the mass of object B. Both objects are moving at the same speed. Which accurately describes how inertia relates to Newton’s second law of motion in this example? Object A has twice the mass of object B. Both objects are moving at the same speed.
A. Object A requires twice the force to stop as Object B.
B. Object A requires one-and-a-half times the force to stop as Object B.
C. Object A requires four times the force to stop as Object B.
Answer:
A. Object A requires twice the force to stop as Object B.
Explanation:
Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
[tex] Acceleration = \frac {Net \; force}{mass} [/tex]
Let's assume the following values;
Mass of object B = 10 kg
Mass of object A = 2 * B = 2 * 10 = 20 kg
Acceleration = 5 m/s²
I. To find the force for B;
[tex] Force = mass * acceleration [/tex]
[tex] Force = 10 * 5 [/tex]
Force B = 50 Newton
II. To find the force for A;
[tex] Force = mass * acceleration [/tex]
[tex] Force = 20 * 5 [/tex]
Force A = 100 Newton
From the calculation, we can deduce that Force A (100 N) is twice or double the value of Force B (50 N).
In conclusion, since object A has twice the mass of object B and both objects are moving at the same speed, object A would require twice the force to stop as Object B.
can someone please help
Answer:
Protein bars
Explanation:
Protein bars contain low-sugar and have high protein content and are therefore good source of nutrient when it is required to take food on-the-go so as to keep the muscle mass the same when it is not possible to take a meal on busy days or before workout
Protein bars can also be taken as part of easy and fast breakfast when travelling so as to maintain a balanced blood sugar level
what is the mystery behind black hole
Well since the gravitational pull is so strong, nothing can escape from it which means we haven't been able to get close enough to really know everything about them. There's no way of knowing what truly goes on inside of them since there is no way out.
Answer:
A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation such as light—can escape from it. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.
Explanation:
b) The relation of the gravitational constant is given as G = If the d mm between two masses is increased, will the value of G increase, decrease or re constant? Write in short.
Answer:
G is universally constant
Explanation:
The Gravitational constant G, is a proportionality constant for all objects in the universe with a value of ; G = 6.67*10^-11 m³kg^-1s^-2.
According to the the Newton's Law of Universal Gravitation ; Every object in the universe attracts one another with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.
F α m1m2 α 1/r²
F α m1m2/r²
F = Gm1m2 / r²
m1 and m2 are the masses ; r = distance between them
G = Gravitational constant.
What is another way to describe the vector below?
"40 feet to the right"
A. 40 feet to the left
B. -40 feet to the left
C. -40 m to the left
D. 40 m to the left
Answer:
-40 feet to the left
Explanation:
opposite of 40 feet from right is -40 to left
Answer:
B. -40 feet to the left
The length of second hand of clock is 14cm, an ant sits on the top of second hand. find the following
i) speed of ant
ii) distance covered by ant in 150 seconds
iii) displacement in 150 seconds
Answer:
i) v = 1.47 cm/s
ii) distance = 219.8 cm
iii) displacement = 28cm
Explanation:
Remember that:
The motion of the tip of the second hand is a circular motion.
For something that rotates with an angular frequency ω, and a radius R, the velocity is given by:
v = ω*R.
i) We know that the second hand of a clock does a complete rotation each 60 seconds.
Then the period is:
T = 60s
And the frequency is the inverse of the period, so:
f = 1/T
f = (1/60s)
And the angular frequency is 2*pi times the normal frequency, thus:
ω = 2*pi*f
ω = 2*pi*(1/60s) = (2*pi/60s)
And the radius will be 14 cm, the velocity of the ant is:
v = (2*pi/60s)*14cm
if we replace pi by 3.14 we get:
v = (2*3.14/60s)*14cm = 1.47 cm/s
ii) The distance covered by the ant in 150 seconds:
Remember that the period of the clock is T = 60s
so in 150 seconds we have:
150s/60s = 2.5 revolutions.
Then the total distance covered is 2.5 times the perimeter of a circle of radius R = 14cm, this is:
distance = (2.5)*2*pi*14cm
= (2.5)*2*3.14*14cm = 219.8 cm
iii) We want to know the displacement, this is, the difference between the final position and the initial position.
In 150 seconds, the ant does 2.5 revolutions.
So the ant will end in the opposite side of the circle where she started (if the ant started when the second hand was at the "3", then the final position is when the second hand is at the "9").
So the displacement will be equal to twice the radius, or the diameter of the circle.
if the radius is 14cm, the diameter is:
2*14cm = 28cm
the displacement is 28cm
What is resistance? Difference between resistance and resistivity.
Answer:
Resistance is a measure of the opposition to current flow in an electrical circuit.All materials resist current flow to some degree. They fall into one of two broad categories: Conductors: Materials that offer very little resistance where electrons can move easily.
a student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g=9.8 m/s^2). what is the ball's speed just before hitting the ground?
Answer:
Explanation:
Use the one-dimensional equation for motion
v = v₀ + at and filling in,
v = 0 + (-9.8)(3.32) so
v = -33 m/s (negative because it is going downwards and upwards is positive).
Pls answer……………………..
2nd egg experienced more impulse
Hope this helps! :)
rocks from volcanoes are used by scientists to study the interior of the earth
TRUE OR FALSE
What is shunt resistance? How does it help in measuring current?
Answer:
it helps me
Explanation:
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....
A 4 mm object is observed by a magnifying lens as 1.25 cm. Calculate the ratio do/di.
Answer:
dₒ/dᵢ = 0.32
Explanation:
From the question given above, the following data were obtained:
Object height (Hₒ) = 4 mm
Image height (Hᵢ) = 1.25 cm
Ratio of object distance (dₒ) to image distance (dᵢ) [dₒ/dᵢ] =?
Next, we shall convert 4 mm to cm. This can be obtained as follow:
10 mm = 1 cm
Therefore,
4 mm = 4 mm × 1 cm / 10 mm
4 mm = 0.4 cm
Next, we shall determine the magnification of the lens. This can be obtained as follow:
Object height (Hₒ) = 4 mm
Image height (Hᵢ) = 1.25 cm
Magnification (M ) =?
M = Hᵢ/Hₒ
M = 1.25 / 0.4
M = 3.125
Finally, we shall determine the ratio of object distance (dₒ) to image distance (dᵢ). This can be obtained as follow:
Magnification (M) = 3.125
Ratio of object distance (dₒ) to image distance (dᵢ) [dₒ/dᵢ] =?
Magnification (M) = image distance (dᵢ) / object distance (dₒ)
M = dᵢ/dₒ
Invert
1/M = dₒ/dᵢ
1/3.125 = dₒ/dᵢ
dₒ/dᵢ = 0.32
Therefore, the ratio of object distance (dₒ) to image distance (dᵢ) [dₒ/dᵢ] is 0.32