A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the speed and direction of the satellite after the crashes. Why would the crash affect the motion of the satellite, and which crash caused a greater change in motion for the satellite?
WILL GIVE BRAINLIEST
Answers
Answer:
The fastest satellite must change orbit
The most massive body (m₁) transfers more momentum to the satellite,
Explanation:
For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants
Most massive body (m1)
initial. Before the crash
p₀₁ = M v + m₁ v₁
after the crash
[tex]p_{f1}[/tex] = M v´ + m₁ v₁´
how momentum is conserved
p₀ = p_{f}
Lighter body (m2)
p₀₂ = M v + m₂ v₂
p_{f2} = M v´ + m₂ v₂´
Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.
We can see that p₀₁> p₀₂
Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.
The fastest satellite must change orbit, thus rotating at a different distance from Earth
Related Questions
I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?
a. 250 kg
c. 10 Kg
b. .002 Kg
d. 45 Kg
Answers
Answer:
a. 250kg I think it's the right answer. hope it helps:)
Answer:
C.10
Explanation:
because when you divide 50 divided by 5 = 10
Use the drop-down menus to complete each sentence.
A plant grows toward a sunny window. This response is an example of
✔ phototropism
.
Sometimes a plant grows around a tree for support. This response is an example of
✔ thigmotropism
.
The roots of a plant grow toward a water source. This response is an example of
✔ hydrotropism
.
The roots of a plant grow down into the soil. This response is an example of
✔ gravitropism
.
the answers are already there
Answers
Answer:
The correct answer is - phototropism, thigmotropism, hydrotropism, and gravitropism in order ( already match correctly).
Explanation:
phototropism is a phenomenon in which plants grow towards the light or sun which is accomplished by the hormone auxin in the cells far from the light.
Thigotropism is a type of plant growth that occurs around the tree to support itself which is a touch stimuli response.
The movement of the plant in the direction of the source of the water is known as hydrotropism. In which stimuli is humidity or the water concentration.
The movement of the plant or roots of the plants towards the soil or earth is known as gravitropism here gravity is the stimuli.
Answer:
Use the drop-down menus to complete each sentence.
A plant grows toward a sunny window. This response is an example of
phototropism
.
Sometimes a plant grows around a tree for support. This response is an example of
thigmotropism
.
The roots of a plant grow toward a water source. This response is an example of
hydrotropism
.
The roots of a plant grow down into the soil. This response is an example of
gravitropism
.
Explanation:
what are the laws of newton
Answers
Answer:
Explanation:
These are the laws of Newton
Answer:
the first law, an object will not change its motion unless a force acts upon it. the 2nd one, the force of an object is equal to its mass times it acceleration. the 3rd one is when 2 objects interact, they apply forces to each other of equal magnitude and opposite direction.
A long wire carries a current 5 A from west to east. A magnetic compass pointing North is placed underneath the wire at a distance of 2 mm. What is the deflection of the compass when it is placed under the wire?
Answers
Answer:
no deflection
Explanation:
current is flowing from west to east. As the magnetic field of a long wire carrying current is circular, its direction will be north below the wire and south above the wire (according to the right hand rule). So, when the compass is placed underneath the wire, it will still point towards the north direction.
A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself.
Answers
Answer:
m = 876.71 kg
Explanation:
This is an exercise of Archimedes' principle, which states that the thrust on a body is equal to the weight of the dislodged liquid
B = ρ g V
therefore the load that the balloon can lift is
B - W_structure - w_load = 0
w_load = B - W_structure
The volume of the balloon is
v = 4/3 π r³
let's substitute
w_carga = rho g 4/3 π r³ - m_structure g
the air density at T = 25ºc is ρ = 1.18 kg / m³
let's calculate
w_load = 1.18 9.8 4/3 π 7.15³ - 930 9.8
w_load = 17705,77 - 9114
w_ load = 8591.77 N
this corresponds to a mass of
w_load = m g
m = w_load / g
m = 8591.77 / 9.8
m = 876.71 kg
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answers
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will ________ because the toy car ____
Answers
Answer:
If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration.
Explanation:
I hope this helped
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car gains more potential energy at the higher starting point.
When a toy car moves along a sloped racetrack, it converts its potential energy (due to its height above the ground) into kinetic energy (energy of motion). The higher the starting height of the racetrack, the more potential energy the toy car possesses initially.
As the toy car moves down the sloped track, it will accelerate due to the force of gravity. The potential energy is converted into kinetic energy, and the car's speed increases. According to the law of conservation of energy, the total mechanical energy (sum of potential and kinetic energy) remains constant as long as no external forces, such as friction, act on the car.
Therefore, if the starting height of the racetrack is increased, the toy car will have more potential energy to start with. As it moves down the track, it will convert this increased potential energy into kinetic energy, resulting in a higher speed compared to when it starts from a lower height.
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15 points.
An object of mass 100 kg is observed to accelerate at a rate of 15
m/s/s. Calculate the force required to produce this acceleration.
Answers
Answer:
its 0.5 for all i beleive
Explanation:
A seated musician plays a C4 note at 262 Hz . How much time Δ does it take for 346 air pressure maxima to pass a stationary listener?
Answers
Answer:
t = 1.32 s
Explanation:
We are given;. Frequency of C4 note; F_c = 262 Hz
In conversions, we know that 1 Hz = 1 cycle/s
Thus, F_c = 262 cycles/s
Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.
346 air pressure maxima denotes that the air pressure maxima is 346 cycles.
Thus, time will be;
t = 346 cycles/262 cycles/s
t = 1.32 s
The time taken for the musical note to pass the stationary listener is 1.32 s.
The given parameters:
frequency of the C4 note, f = 262 Hzair pressure maximum, n = 346The frequency of a sound wave is defined as the number of cycles completed per second by the wave.
[tex]F = \frac{n}{t}[/tex]
where;
t is the time to compete the maximum cycleThe time taken for the musical note to pass the stationary listener is calculated as follows;
[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]
Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.
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