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2 years ago

A speck of dust on a spinning dvd has a centripetal accelera- tion of 20 m/s2.

2 answers:

alexandr1967 [171]2 years ago

7 0

a) 40 m/s^2 b) 80 m/s^2 The formula for centripetal force is F = mv^2/r With that in mind, let's look at the questions. a. what is the acceleration of a different speck of dust that is twice as far from the center of the disk? This basically asks "What changes is r is doubled?" and looking at the formula, the effects of doubling r are two fold. First, velocity is directly related to radius, so the term v^2 is multiplied by 4 which increases the force by a factor of 4. But that's not all. The r term in the divisor is also doubled which will cause the force to be divided by a factor of 2, so the force is halved. So one half of 4 is 2. So the total force is doubled. So 20 m/s^2 * 2 = 40 m/s^2 b. what would be the acceleration of the first speck of dust if the disk's angular velocity was doubled? This is more straight forward. The v term in the formula for force is doubled. So the value of v^2 will be quadrupled. Which in turn causes the overall force to quadruple. So 20 m/s^2 * 4 = 80 m/s^2

ArbitrLikvidat [17]2 years ago

6 0

The centripetal acceleration of an object spinning in a circle is
a = v²/r = 20 m/s²
where
r = the distance of the object from the center of the circle
v = the tangential velocity of the object

The centripetal acceleration is also given by
a = rω² = 20 m/s²
where
ω = the angular velocity.

Part a.
If the radius is doubled while the angular velocity remains constant, then
the centripetal acceleration will double.
a = (2r)ω² = 2*20 = 40 m/s²

Answer: 40 m/s²

Part b.
If the angular velocity is doubled, then
a = r(2ω)² = 4rω² = 4*20 = 80 m/s²

Answer: 80 m/s²

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The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

julia-pushkina [17]

Answer: They are all true

a. The tension in the rope is everywhere the same.

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d. The forces exerted on the two objects by the rope must be in the direction of the rope.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)

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2 years ago

89. An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes a

ddd [48]

Explanation:

Given that,

Initial speed of the electron, $u=4\times 10^5\ m/s$

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, $a=6\times 10^{12}\ m/s^2$

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

$v^2=u^2+2as$

$v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05$

v = 871779.788 m/s

$v=8.71\times 10^5\ m/s$

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

$t=\dfrac{v-u}{a}$

$t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}$

$t=7.85\times 10^{-8}\ s$

Hence, this is the required solution.

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2 years ago

Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288

Tomtit [17]

Answer 
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football.

7 0

2 years ago

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Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1

abruzzese [7]

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

3 0

2 years ago

A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?

andreev551 [17]

Hello!

We have the following data:

Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

$P = \frac{E}{T}$

Solving:

$P = \frac{E}{T}$

$P = \frac{E}{T} \to T = \frac{E}{P}$

$T = \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}$

$\boxed{T = 3000\:seconds}$

How many minutes can it run for? (Let's convert in minutes)

1 minute --------- 60 seconds
y minute --------- 3000 seconds

$\frac{1}{y} = \frac{60}{3000}$

Product of extremes equals product of means

$60*y = 1*3000$

$60y = 3000$

$y = \frac{3000}{60}$

$\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark$

I hope this helps! =)

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2 years ago

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