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Trava [24]
2 years ago
11

A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of

the loop in order to remain in contact with the loop
Physics
1 answer:
otez555 [7]2 years ago
4 0
<span>3.834 m/s. In this problem we need to have a centripetal force that is at least as great as the gravitational attraction the object has. The equation for centripetal force is F = mv^2/r and the equation for gravitational attraction is F = ma Since m is the same in both cases, we can cancel it out and then set the equations equal to each other, so a = v^2/r Substitute the known values (radius is diameter/2) and solve for v 9.8 m/s^2 <= v^2/1.5 m 14.7 m^2/s^2 <= v^2 3.834057903 m/s <= v So the minimum velocity needed is 3.834 m/s.</span>
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An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spr
neonofarm [45]

Answer:

2.63 Hz

Explanation:

m = mass of the object = 8.0 kg

x = stretch in the spring = 3.6 cm = 0.036 m

k = spring constant of the spring

using equilibrium of force

Spring force = weight of object

k x = m g

k (0.036) = (8) (9.8)

k = 2177.78 N/m

frequency of oscillation is given as

f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}

f = \frac{1}{2\pi }\sqrt{\frac{2177.78}{8}}

f =  2.63 Hz

4 0
2 years ago
Read 2 more answers
Complete the passage to identify potential and kinetic energy. A rock resting on the top of a hill has energy, while a rock roll
kipiarov [429]
A rock resting on the top of a hill has POTENTIAL energy, while a rock
rolling down a hill has KINETIC energy.
8 0
2 years ago
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A body covers a semicircle of radius 7cm in 5s .find its linear speed
choli [55]

Ok so we are given the radius of 7cm and time of 5 seconds.

From the data we got we can calculate speed, frequency, perimeter and area of the semicircle.

Let's start with perimeter.

We know that perimeter of circle is 2\pi r so the perimeter of semicircle is \dfrac{2\pi r}{2} or simply \pi r

So the perimeter is equal to:

\pi r=\pi\cdot7\approx\boxed{22cm}

So this is the length of a curve or let's say the distance.

Now let's look at the linear speed s=\dfrac{d}{t} where d is distance and t time.

We know the distance and we know the time.

So let's calculate it.

s=\dfrac{d}{t}=\dfrac{22}{5}=\boxed{4.4\dfrac{cm}{s}}

Hope this helps.

r3t40

8 0
2 years ago
4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity
Slav-nsk [51]

Answer:

w_f= 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

w_f^2-w_i^2=2aS

where w_f is the final angular velocity, w_i the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

w_f^2-(0)^2=2(4)(62.83)

Finally, solving for w_f:

w_f= 22.41rad/s

8 0
2 years ago
An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of
Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 /  .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B)  Rest length of the rocket = length of platform = 90 m

C ) length of platform  as viewed by moving observer =

\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }

= \frac{90}{\sqrt{1-\frac{0.92}{1 } } }

= 321 m

D )  For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }

8.75 x 10⁻⁸ s .

8 0
2 years ago
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