Answer:
- Distance is a scalar quantity, defined as the total amount of space covered by an object while moving between the final position and the initial position. Therefore, it depends on the path the object has taken: the distance will be minimum if the object has travelled in a straight line, while it will be larger if the object has taken a non-straight path.
- Displacement is a vector quantity, whose magnitude is equal to the distance (measured in a straight line) between the final position and the initial position of the object. Therefore, the displacement does NOT depend on the path taken, but only on the initial and final point of the motion.
If the object has travelled in a straight path, then the displacement is equal to the distance. In all other cases, the distance is always larger than the displacement.
A particular case is when an object travel in a circular motion. Assuming the object completes one full circle, we have:
- The distance is the circumference of the circle
- The displacement is zero, because the final point corresponds to the initial point
The tension in the supporting cable when the cab originally moves downward is 18422.4 N
What is tension?
Tension is described as the pulling force by the means of a three-dimensional object.
Tension might also be described as the action-reaction pair of forces acting at each end of said elements.
Here,
m =combined mass = 1600 kg
s = Displacement of the elevator = 42 m
g = Acceleration due to gravity = 9.81 m/s²
u = Initial velocity = 12 m/s
v = Final velocity = 0
According to the equation of motion:
0 - 12^2 = 2*a*42
a = - 144 / 84
a = - 1.714 m/s^2
Now let's write the equation of the forces acting on the elevator. Taking upward as positive direction:
T-mg = ma
T = m(g-a)
T = 1600 ( 9.8-(-1.74))
T=18422.4 N
Hence,
The tension in the supporting cable when the cab, originally moving downward is 18422.4 N
Learn more about tension here:
<u>brainly.com/question/13772148</u>
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Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
Multiply each side
by 0.09 m³ : (4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg .
Force of gravity = (mass) x (acceleration of gravity)
= (360 kg) x (9.8 m/s²)
= (360 x 9.8) kg-m/s²
= 3,528 newtons .
That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).
Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.
The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.
So while it's in the water, the block seems to weigh
(3,528 - 882) = 2,646 newtons (about 595.2 pounds) .
But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
The average distance between the planet and the star is:
R=9.36*10^11 m
Orbital velocity v=√{(G*M)/R},
G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,
M = mass of the star
R =distance from the planet to the star.
v=ωR, with ω as the angular velocity and R the radius
ωR=√{(G*M)/R},
ω=2π/T,
T = orbital period of the planet
To get R we write the formula by making R the subject of the equation
(2π/T)*R=√{(G*M)/R}
{(2π/T)*R}²=[√{(G*M)/R}]²,
(4π²/T²)*R²=(G*M)/R,
(4π²/T²)*R³=G*M,
R³=(G*M*T²)/4π²,
R=∛{(G*M*T²)/4π²},
Substitute values
R=9.36*10^11 m
As was already said, Earth is located roughly 150 million kilometres (93 million miles) from the Sun on average. It is 1 AU. Mars is on our fictitious football field's three-yard line. On average, the distance between the Sun and the red planet is around 142 million miles (228 million kilometres).
Learn more about average distance:
brainly.com/question/18366547
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The complete question is ''If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 10^30 kg, what is the average distance between the planet and the star? 1.43 × 10^9 m 9.36 × 10^11 m 5.42 × 10^13 m 9.06 × 10^17 m''.
The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>
