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mr_godi [17]
3 years ago
9

How do you calculate centripetal acceleration?

Physics
1 answer:
Delvig [45]3 years ago
8 0

A particle has centripetal acceleration whenever it's a making a turn of radius R. If the particle is moving at a constant tangential speed v throughout the turn, then the magnitude of centripetal acceleration is

v²/R

If the particle is following a uniformly circular path, then it moves in a circle of radius R and travels a distance equal to its circumference, 2πR. Let T be the time it takes to complete one such loop. Then the entire circle is traversed with speed v = 2πR/T, so that the centripetal acceleration is also given by

v²/R = (2πR/T)²/R = 4π²R/T²

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An object has an acceleration of 12.0 m/s/s. The mass of the object is doubled while the net force on the object is held constan
Aloiza [94]

Answer:

The net force acting on the object is doubled while the mass of the object is held constant. What will be the new acceleration? An object has an acceleration of 12.0 m/s^2. The net force acting on the object is halved (decreased to one half its original value) while the mass of the object is held constant.

7 0
3 years ago
A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above
vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft

Earth

s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft

Differentiating the first equation with respect to time we get

\frac{ds}{dt}=96-12t

Equating with zero

0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s

Differentiating the second equation with respect to time we get

\frac{ds}{dt}=96-32t

Equating with zero

0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s

Applying the time taken to the above equations, we get

s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384

s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

6 0
3 years ago
Some curious students hold a rolling race by rolling four items down a steep hill. The four items are a solid homogeneous sphere
Papessa [141]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

8 0
3 years ago
A force F→=(cx-3.00x2)iˆ acts on a particle as the particle moves along an x axis, with F→ in newtons, x in meters, and c a cons
DerKrebs [107]

Answer:

Explanation:

Work done = ∫Fdx

= ∫(cx-3.00x²)   dx

[ c x² / 2 - 3 x³ / 3 ]₀²

= change in kinetic energy

= 11-20

= - 9 J

[ c x² / 2 -  x³   ]₀² = - 9

c x 2² / 2 - 2³ = -9

2c - 8 = -9

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c = - 1/2

6 0
4 years ago
Read 2 more answers
8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included  with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, T_P = 30 N

The tension in rope Q, T_Q = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, \Sigma F_y = T_P_y + T_Q_y + W = 0

∴ W = -(T_P_y + T_Q_y)

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, \Sigma F_x = T_P_x + T_Q_x  = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

3 0
3 years ago
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