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3 years ago

An object is spun around in circular motion such that its frequency is 500 Hz.

Physics

1 answer:

Marrrta [24]3 years ago

7 0

Explanation:

Given parameters:

Frequency = 500Hz

Unknown:

a. Period of rotation

b. Time required to complete 7 rotation

Solution:

The period of rotation is the time taken to complete a turn.

P = $\frac{t}{n}$

P is the period

t is the time

n is the number of turns

Period is the inverse of frequency;

P = $\frac{1}{500}$ = 0.002s

Time required to complete 7 rotation;

P = $\frac{t}{n}$

t = P x n

t is the time

P is the period

n is the number of rotations

t = 0.002 x 7 = 0.014s

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A battery charger can produce 3A at 12 Volt and charges a battery fer 2 hr. Calculate work in KJ.

Oksi-84 [34.3K]

Answer: 259.2 KJ

Explanation:

The formula calculate work don in a circuit is given by :-

$W=QV$ , where Q is charge and V is the potential difference.

The formula to calculate charge in circuit :-

$Q=It$ , where I is current and t is time.

Given : Current : $I=3A$

Potential difference : $V=12\ V$

Time : $t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}$

Now, $Q=3(7200)=21,600\ C$

Then, $W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}$

Hence, the work done = 259.2 KJ

4 0

3 years ago

(a) Calculate the buoyant force on a 2.00-L helium balloon.

iragen [17]

Answer:

0.0239364 N

0.0057879 N

Explanation:

$\rho$ = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

$F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N$

The buoyant force is 0.0239364 N

Net vertical force is given by

$F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N$

The net vertical force is 0.0057879 N

6 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

Star X has an apparent magnitude of 1. Star Y has an apparent magnitude of 4. Both stars are in the same star cluster. Which sta

sashaice [31]

Answer:

Explanation:

From the given information:

Since both stars are in the same cluster, the magnitude and luminosity relationship can be calculated as:

$m_1 - m_2 = -2.5 log _{10} (\dfrac{L_1}{L_2})$

Given that;

m_1 = 1 and

m_2 = 4

Therefore,

$1 - 4 = -2.5 log _{10} ( \dfrac{L_1}{L_2})$

$3 = -2.5 log _{10} ( \dfrac{L_1}{L_2})$

Making $\dfrac{L_1}{L_2}$ the subject of the formula:

$\implies \dfrac{L_1}{L_2}= 10^{(\dfrac{3}{2.5})}$

=15.84

≅ 16

Hence, we can conclude that star X is more luminous by a factor of 16

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3 years ago

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