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Tasya [4]
2 years ago
13

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?
Physics
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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It's nighttime, and you ve dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90 m above the ed
yanalaym [24]

Answer:

The distance of the goggle from the edge is 5.30 m

Explanation:

Given:

The depth of pool (d) = 3.2 m

let 'i' be the angle of incidence

thus,

i = tan^{-1}(\frac{2.2}{0.90})

i = 67.75°

Now, Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

where,

r is the angle of refraction

n₁ is the refractive index of medium 1 = 1 for air

n₂ is the refractive index of medium 1 = 1.33 for water

now,

1 × sin 67.75° = 1.33 × sin(r)

or

r = 44.09°

Now,  

the distance of googles = 2.2 + d×tan(r)  = 2.2 + (3.2 × tan(44.09°) = 5.30 m

Hence, <u>the distance of the goggle from the edge is 5.30 m</u>

5 0
3 years ago
C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-
mylen [45]

Answer:

C-14 has two more neutrons.

Explanation:

4 0
3 years ago
Read 2 more answers
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
2 years ago
What<br> must always be<br> included on the<br> graph
Nookie1986 [14]
Clearly visible data points and appropriate labels on each access that include units
4 0
3 years ago
Does anyone know what the answers are?
baherus [9]

Answer:

Option (C) is the answer

Explanation:

may be it is possible if that we stand so far

4 0
3 years ago
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