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4vir4ik [10]
2 years ago
13

Assuming motion is on a straight path, what would result in two positive components of a vector?

Physics
2 answers:
Aliun [14]2 years ago
3 0
Answer: option <span>A) a train moving north of east at an angle of 25°
</span>

Explanation:

1) You need to choose your axis. In this case North is vertical and positive, South is vertical and negative, East is horizontal and positive, and West is horizontal and negative.

2) The vector with the two positive components is a vector in the first quadrant (North and East). That is what North of East 25° means.

3) Regarding the other options:

<span>B) a bus moving North of East at an angle of 95°: since the angle is greater tnan 90° the vector is in the second quadrant: its horizontal component is negative.
</span>
<span /><span>
</span><span> C) a boat moving South of West at an angle of 40°: this is in the third quadrant: the two components are negative.</span>
<span /><span>
</span><span> D) a car moving south of west at an angle of 10°: as in the option C), this is in the third quadrant: the two components are negative.
</span>

avanturin [10]2 years ago
3 0

A.

Two positive components means that the vector should lie in the first quadrant and not be parallel to the x or y axis; the only option that satisfies this is A.

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A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. Long
tino4ka555 [31]

Answer:

A) equal to the battery's terminal voltage.

Explanation:

When the capacitor is fully charged after long hours of charging , its  potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and  current becoming zero . There is no charging current when the capacitor is fully charged .

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3 years ago
In the equation for Ohm’s law what does I stand for?
Ket [755]
A) current 

(I is always current in electricity)
3 0
3 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
2 years ago
A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Anit [1.1K]

Answer: The gravitational force Fg exerted on the orbit by the planet is Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

Explanation:

Gravitational Force Fg = GMm/r2----1

Where G is gravitational constant

M Mass of the planet, m mass of the orbit and r is the distance between the masses.

Since the circular orbit move around the planet, it means they do not touch each other.

The distance between two points on the circumference of the two massesb is given by d, while the distance from the radius of each mass to the circumferences are R1 and R2 from the question.

Total distance r= (R1 + d + R2)^2---2

Recall, density rho =

Mass M/Volume V

Hence, mass of planet = rho × V

But volume of a sphere is 4/3πr3

Therefore,

Mass M of planet = rho × 4/3πr3

=4/3πr3rho in kg

From equation 1 and 2

Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

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For plants that can genetically self pollinate, but would prefer not to, they can avoid this by having their pistil and pollen/stamens mature at different times. If the stamens mature first, the pollen will be dispersed by animals or wind or whatever dispersal mechanism it relies on. Then by the time the pistil is ready to be pollinated, there is no pollen left in that flower to land on the stigma.

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