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nexus9112 [7]
3 years ago
5

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

that the ball in her hand moves in a circle. In one instance, the radius of the circle is 0.610 m. At one point on this circle, the ball has an angular acceleration of 67.6 rad/s2 and an angular speed of 17.0 rad/s. (a) Find the magnitude of the total acceleration (centripetal plus tangential) of the ball. (b) Determine the angle of the total acceleration relative to the radial direction.
Physics
1 answer:
Pepsi [2]3 years ago
4 0

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

a_r=\omega^2R

Plug in the given values and solve for a_r. This gives,

a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2

Now, tangential acceleration is given by the formula:

a_t=R\alpha

Plug in the given values and solve for a_t. This gives,

a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

a_{Total}=\sqrt{(a_r)^2+(a_t)^2}

Plug in the given values and solve for total acceleration, a_{Total}. This gives,

a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

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a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin
Sveta_85 [38]

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

6 0
4 years ago
1pt A cannon fires a 5-kg ball horizontally from a
Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0
3 years ago
Blake is setting up his tent at a renaissance fair. If the tent is 8 feet tall and the tether can be staked no more than 2 feet
Mariana [72]
The stake, height and tether length of the tent form a right angle triangle where the tether length is the hypotenuse.
Applying Pythagoras theorem:
length² = height² + (stake distance)²
length = √(8² + 2²)
length = 8.5 feet
3 0
3 years ago
A ball hits a wall horizontally at 6m/s and rebounces at 4.4m/s the ball is in contact with wall for 0.04 sec. what is the accel
Svetlanka [38]

Answer:

Acceleration (a) = 40 m/s²

Explanation:

Given:

Initial velocity (u) = 6 m/s

Final velocity (v) = 4.4 m/s

Time taken (t) = 0.04sec

Find:

Acceleration (a) = ?

Computation:

We know that,

⇒ v = u + at

⇒ a = (v - u) / t

⇒ Acceleration (a) = (4.4 - 6) / 0.04

⇒ Acceleration (a) = (-1.6) / 0.04

Acceleration (a) = 40 m/s²

8 0
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A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This fo
Gwar [14]

Answer:

m = 95000 kg

Explanation:

Given that,

Net force acting on the house, F = 2850 N

Initial speed, u = 0

Final speed, v = 15 cm/s = 0.15 m/s

We need to find the mass of the house. Let the mass be m. We know that the net force is given by :

F = ma

Where

a is the acceleration of the house.

So,

F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg

So, the mass of the house is equal to 95000 kg.

3 0
3 years ago
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