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gavmur [86]
3 years ago
6

Under what conditions can an object travel a certain distance and yet it's resultant displacement be zero?

Physics
1 answer:
Veseljchak [2.6K]3 years ago
4 0
If a person returns to the position he started again. like moving in a circle track
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A rope is pulling a sled is sloped at 25 degrees upward at the point of contact. The tension in the rope is 40.0N. What is the h
pshichka [43]
T = 40.0 N

angle = 25°

Trigonometric ratios:

sin(25°) = vertical component of the force / force

cos(25°) = horizontal component of the force / force

tan (25°) = vertical compoent of the force / horizontal component of the force.

From cos(25°) you can find the horizontal component of the force:

horizontal component of the force = force * cos(25°)

The force is the tension, 40.0 N.

horizontal component of the force = 40.0 N * cos (25°) = 36.25 N

Answer: 36.25 N
7 0
4 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

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8 0
2 years ago
A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from
Rzqust [24]

The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

Learn more about brainly.com/question/874205 here:

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7 0
2 years ago
Read 2 more answers
Why does the north poll of Jupiter form hexagonal shapes in the clouds of Jupiter. Trying to see if people got brains
inn [45]

Explanation:

They believe that vortexes occur at the planet's north pole because of atmospheric flows deep within the gas giant, and that these vortexes pinch an intense horizontal jet near the equator—which is what warps the storm into a hexagon

8 0
4 years ago
(b) If the object is at 330 feet and its instantaneous velocity is 3 feet per minute at 30 minutes, what is the approximate posi
ELEN [110]

Answer:

The final position is 36 feet.

Explanation:

initial position, d = 330 feet

speed, v = 3 feet per minute

time, t = 30 minute

now the time is 32 minute

time interval = 2 minute

So, the distance in 2 minutes is

d' = 2 x 3 = 6 feet

So, the final position is

D = 30 + 6 = 36 feet

8 0
4 years ago
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