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ddd [48]
3 years ago
9

A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t

he corner?
Physics
1 answer:
damaskus [11]3 years ago
3 0
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,

                  a = v² / r

where a is the acceleration, v is the velocity, and r is the radius.

Substituting the known values from the given above,

             4.4 m/s² = (15 m/s)² / r

The value of r from the equation is 51.14 m.

Answer: 51.14 m
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You are sitting in a car that is travelling at a constant 10 m/s. How fast is your drink, sitting in the cup holder, traveling r
Ksenya-84 [330]

Here, we are required to determine how fast is you drink, sitting in the cup holder, travelling relative to the car.

  • The speed of the drink, sitting in the cup holder, relative to the car is; 0m/s

From the laws of relative motion,

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)</em>

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)</em>

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)when object A and Object B are travelling with speed a and b respectively in the same direction, where speed a = speed b, then the speed of object A relative to object B is; zero(0).</em>

Evidently, the scenario in the question is similar to the third scenario above. The cup, sitting in the cup holder is travelling with the car at the same constant speed 10m/s.

Therefore, the speed of the drink relative to the car is zero(0).

Read more:

brainly.com/question/20549055

7 0
1 year ago
How you can speed up the dissolving process when preparing juice from frozen concentrate
Katyanochek1 [597]

If you stir the juice it increases the surface area.

8 0
2 years ago
1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of
laiz [17]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

  • constant velocity of the elevator, u₁ = 10 m/s
  • initial velocity of the ball, u₂ = 20 m/s
  • height of the boy above the elevator floor, h₁ = 2 m
  • height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\  gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s

To learn more about projectile calculations please visit: brainly.com/question/14083704

6 0
2 years ago
How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?
Nady [450]

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations.  We are going to use vf = vi + at.  Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.  

5 0
3 years ago
Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ ​ 3 2 4 ​ ⎦ ⎥ ⎤ ​ km away from the origin with respect to so
Anettt [7]

Answer:

[1, 6, -2]

Explanation:

Given the following :

Initial Position of spaceship : [3 2 4] km

Velocity of spaceship : [-1 2 - 3] km/hr

Location of ship after two hours have passed :

Distance moved by spaceship :

Velocity × time

[-1 2 -3] × 2 = [-2 4 -6]

Location of ship after two hours :

Initial position + distance moved

[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]

= [3-2, 2+4, 4-6] = [1, 6, -2]

4 0
3 years ago
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