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3 years ago

A girl running around a circular track of radius 200 m completes one

Physics

1 answer:

kondor19780726 [428]3 years ago

4 0

Answer:

v = 2 m/s

Explanation:

First, we find the angular velocity of the girl by using the following formula:

$Angular\ Velocity =\omega= \frac{Angular\ Distance\ Covered}{Time\ Taken}\\\\\omega = \frac{1\ revolution}{10\ min}\frac{1\ min}{60\ s}\frac{2\pi\ rad}{1\ revolution}\\\\\omega = 0.01\ rad/s$

Now, for the average velocity of the girl we can use the following formula:

$Average\ Velocity = V = r\omega\\V = (200\ m)(0.01\ rad/s)\\$

where,

r = radius = 200 m

therefore,

v = 2 m/s

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An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

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3 years ago

A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a

sleet_krkn [62]

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m

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2 years ago

Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro

alex41 [277]

They were formed in the nuclear fusion reaction inside older stars.

As a star burns, fusion reactions inside its core create heavier elements. Those materials are released when the star dies of old age in an explosion.

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3 years ago

Image caught on Screen is called

bixtya [17]

Answer:

<h2>Virtual image</h2>

Explanation:

<h3>Virtual image can be caught on a screen</h3>

hope this helps you.

will give the brainliest!

follow ~Hi1315~

4 0

3 years ago

Read 2 more answers

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila

Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

$\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}$

substituting the values in the above question we get

$\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}$

$\frac{A_1}{A_2} }$ =0.3659

6 0

3 years ago