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2 years ago

Jennifer's cat is stuck in a

Physics

1 answer:

Lelu [443]2 years ago

5 0

The speed with which Jennifer climb up the tree is 0.45 m/s

The distance is 10m

Time is 22 seconds

Required speed is 10/22=0.45 m/s

Distance is the total length travelled by an object.
Speed is distance upon time.

It is a scalar quantity and doesn't has magnitude and direction.

Distance traveled is the total length of the path traveled between two positions
Distance is the total movement of an object without any regard to direction.
Distance is the total movement of an object without any regard to direction.

To learn more about distance visit at :

brainly.com/question/15256256

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A coin rests 11.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface i

9966 [12]

Answer:

5.5 rad/s

Explanation:

The friction between the coin and the turntable provides the centripetal force that keeps the coin in circular motion. Therefore, we can write:

$\mu mg = m\omega^2 r$

where

$\mu = 0.340$ is the coefficient of friction

m is the mass of the coin

$g=9.8 m/s^2$ is the acceleration of gravity

$\omega$ is the angular speed

r is the distance of the coin from the centre of rotation

In this problem,

r = 11.0 cm = 0.11 m

The coin starts to slip when the centripetal force becomes larger than the maximum frictional force:

$m\omega^2 r > \mu m g$

Solving for $\omega$ , we find the angular speed at which this happens:

$\omega = \sqrt{\frac{\mu g}{r}}=\sqrt{\frac{(0.340)(9.8)}{0.11}}=5.5 rad/s$

5 0

3 years ago

What year did Clara Baer invent Nukem/Newcomb ball?

Semenov [28]

Answer:

1895

Explanation: it was invented in 1895 by Clara Berra a physical education instructor at Sophie Newcomb College.

7 0

3 years ago

A production machine that consists of three components connected in series. The first component follows Weibull probability dist

nika2105 [10]

Answer:

The correct solution is "0.66104".

Explanation:

Given:

Component 1,

$\beta=1.3$

$\gamma=24,000 \ hr$

Component 2,

$\beta = 1.9$

$\gamma=18,000 \ hr$

Component 3,

$MTTF=48,000 \ hr$

Now,

At t = 6000 hr, the system reliability will be:

⇒ $Rs(t=6000)=\frac{3}{\pi} R_1\times R_2\times R_3$

$=[e^{-(\frac{6000}{24000} )^{1.3}}]\times [e^{-(\frac{6000}{18000} )^{1.9}}]\times [e^{-(\frac{6000}{48000} )}]$

$=0.66104$

Thus the above is the correct solution.

3 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs.

love history [14]

Answer:

Georges Lemaître - theorized that the universe had a beginning

Albert Einstein - huit a model of the inverse haced on relativity

Galileo Galilei - was the first to use a telescope to observe the planets

Edwin Hubble - showed that the universe was expanding

Henrietta Swan Leavitt - demonstrated a method to calculate the distance of celestial bodies

is my choices report if wrong

5 0

3 years ago

A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is

Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d= $\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m$

Wavelength of light, $\lambda=510 nm=510\times 10^{-9} m$

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

$sin\theta=\frac{m\lambda}{d}$

For first bright fringe, =1

$sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305$

$\theta=sin^{-1}(0.305)=17.76^{\circ}$

The distance between two m=1 fringes

$x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m$

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

$sin\theta=1$

$sin\theta=\frac{m\lambda}{d}$

Substitute the values

$1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}$

$m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3$

Maximum number of bright fringes on the scree= $2m+1=2(3)+1=7$

8 0

4 years ago