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3 years ago

A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2m/s to 4 m/s?

2 answers:

6 0

Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s

In short, Your Answer would be Option C

Hope this helps!

olganol [36]3 years ago

5 0

Answer: The initial momentum is 12 kg m/s, and the final momentum is 24 kg m/s

Explanation:

Momentum is defined as the force that keeps the object moving. It is also defined as the product of mass and velocity of an object.

Mathematically,

$p=m\times v$ .......(1)

where,

p = momentum of the bowling ball

m = mass of the bowling ball

v = velocity of the bowling ball

Calculating the initial momentum:

We are given:

$m=6kg\\u=\text{Initial speed}=2m/s$

Putting values in equation 1, we get:

$p=6kg\times 2m/s\\\\p=12kg.m/s$

Calculating the final momentum:

We are given:

$m=6kg\\u=\text{Final speed}=4m/s$

Putting values in equation 1, we get:

$p=6kg\times 4m/s\\\\p=24kg.m/s$

Hence, the initial momentum is 12 kg m/s, and the final momentum is 24 kg m/s

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Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

$m=\dfrac{F}{g}$

$m=\dfrac{160\ lb}{32\ ft/s^2}$

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

$m=\dfrac{F}{g}$

$m=\dfrac{1.9\ lb}{32\ ft/s^2}$

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

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(b) A 22 g song sparrow, m = 22 g = 0.022 kg

$W=0.022\ kg\times 32\ ft/s^2=0.704\ lb$

Hence, this is the required solution.

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2 years ago

A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frict

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B it’s Intail velocity

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2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

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a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

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b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

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$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

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Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

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Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

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3 years ago

With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as

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Answer:

|Δf| = 37.3 kHz

Explanation:

given,

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speed of the sound = 1500 m/s

frequency = 7 MHz

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$\delta f = \pm 2 f_0 (\dfrac{V}{C})$

$\delta f = \pm 2\times 7 (\dfrac{4}{1500})$

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|Δf| = 37.3 kHz

hence, frequency shift between the opening and closing valve is 37.3 kHz

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