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2 years ago

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A 0.203-kg plastic ball moves with a velocity of 0.30 m/s. it collides with a second plastic ball of mass 0.092 kg, which is mov

ing along the same line at a speed of 0.10 m/s. after the collision, both balls continue moving in the same, original direction. the speed of the 0.092-kg ball is 0.26 m/s. what is the new velocity of the 0.203-kg ball?

1 answer:

dolphi86 [110]2 years ago

7 0

<span><span>A 0.200 kg plastic ball moves with a velocity of0.30 m</span>s<span>A 0.205-kg plastic ball moves with a velocity of0.30 m</span><span>A 0.199 kg plastic ball moves with a velocity of0.30 m</span><span>A 0.204-kg plastic ball moves with a velocity of0.30 m</span><span>A 100 g ball moving to the right at 4.0 m</span>s collides<span>have less momentum if the velocities</span><span>the same</span><span>A ball with a momentum of 4.0 kg•<span>m</span></span></span>

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A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

2 years ago

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.5 rad/s. Find the angular displ

zavuch27 [327]

Answer: angular displacement in rad = 3038.45 rad

angular displacement in rev = 483.589 rev

Explanation: mathematically

Angular velocity = angular displacement / time taken.

Angular velocity = 33.5 rad/s, time taken = 90.7s

33.5 = angular displacement /90.7

Angular displacement = 33.5 * 90.7 = 3038.45 rad

But 1 rev =2π

Hence 3038.45 rad to rev is

3038.45/2π = 483.599 rev

7 0

3 years ago

Read 2 more answers

A potential difference of 24 V is applied to a 150-ohm resistor. How much current flows through the resistor?

Lady_Fox [76]

Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm

Potential Difference (v) = 24 V

Current (I) = ?

V = IR

24 = I × 150

I = 24/150

I = 0.16 ampere

hope it helps!

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2 years ago

Find the frequency of a wave of wavelength 2.5m and speed 400 m/s

Sedaia [141]

Answer:

The frequency of wave is 160Hz.

Explanation:

Given that the formula of speed is V = f×λ where V represents speed, f is frequency and λ is wavelength.

So first thing, you have to make frequency the subject by dividing wavelength on both sides :

$v = f \times λ \:$

$v \div λ = f \times λ \div λ$

$f = \frac{v}{λ}$

Next you have to substitute the value of v and f into the formula :

Let λ = 2.5m,

Let v = 400m/s,

$f = \frac{400}{2.5}$

$f = 160Hz$

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2 years ago

As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0

Alika [10]

velocity of the physics instructor with respect to bus

$v = 6 m/s$

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$a = 2 m/s^2$

acceleration of instructor with respect to bus is given as

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now the maximum distance that instructor will move with respect to bus is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 6^2 = 2(-2)(d)$

$-36 = - 4 d$

$d = 9 m$

so the position of the instructor with respect to door is exceed by

$\delta x = 9 - 6 = 3 m$

so it will be moved maximum by 3 m distance

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3 years ago