Ask question

Ask question

All categories

Ostrovityanka [42]

3 years ago

12

As the mass of an object in uniform circular motion increases, what happens to the centripetal force required to keep it moving

at the same speed?

1 answer:

Dmitriy789 [7]3 years ago

7 0

Answer:

It increases, because the centripetal force is directly proportional to the mass of the rotating body.

You might be interested in

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

Llana [10]

Answer:

$K_e_q=22.75878093\frac{N}{m}$

$f=1.363684118Hz$

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

$\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}$

Replacing the data provided:

$\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}$

$K_e_q=22.75878093\frac{N}{m}$

Finally, to calculate the frequency of oscillation we use this:

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

Replacing m and k:

$f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz$

4 0

3 years ago

A boater is floating 1060 meters down a river. The person takes 40 minutes. At what speed is the boat moving?

Mashutka [201]

26.5, I’m not sure if it’s right but 1060/40 gets that

7 0

3 years ago

1)Which statement best describes energy in an open system?

zimovet [89]

B an open system flow both

6 0

3 years ago

Read 2 more answers

An amateur astronomer looks at the moon through a telescope with a 15-cm-diameter objective. What is the minimum separation betw

Lana71 [14]

Answer:

y = 128.0 km

Explanation:

The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains

the diffraction equation for the first minimum is

a sin θ = λ

In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.

θ = 1.22 λ / d

d = 15 cm

to find the distance we can use trigonometry

tan θ = y / L

tan θ = sin θ / cos θ = θ

substituting

y / L = λ / d

y = L λ /d

let's calculate

y = 384 10⁸ 500 10⁻⁹ / 0.15

y = 1.28 10⁵ m

Let's reduce to km

y = 1.28 10⁵ m (1km / 10³ m)

y = 128.0 km

the correct answer is 120 km away

5 0

3 years ago

The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/

nata0808 [166]

Answer:

The tension on the string is 2.353 N.

Explanation:

Given;

the speed of sound in air, v₀ = 343 m/s

then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s

mass per unit length, m/l = μ = 0.002 kg/m

The speed of sound on the string is given as;

$v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu$

where;

T is the tension on the string

T = (34.3)²(0.002)

T = 2.353 N

Therefore, the tension on the string is 2.353 N.

3 0

3 years ago