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3 years ago

7

A 100 kg cart goes around the inside of a vertical loop of a roller coaster. The radius of the loop is 3 m and the cart moves at

a speed of 6 m/s at the top. The force exerted by the track on the cart at the top of the loop is

2 answers:

Inga [223]3 years ago

6 0

Answer:

220N

Explanation:

Weight of the cart = w = 100kg

Radius = r = 3m

Velocity = v = 6m/s

9 = 9.8m/s²

F = ?

Since the cart is at the top of inner loop both the average normal force (N) and the weight (w=mg) of the cart act downwards, hence they both add up. Considering the centrifugal force (F=mv²/r) as well, we get,

∑F=ma=mv²/r

N+mg=mv²/r

N=m(v²/r-g)

N=100[(6)²/3-9.8]

N=100[12-9.8]

Net Force=220N

So, the force exerted by the track on the cart at the top of the loop is 220N

zepelin [54]3 years ago

3 0

Answer:

220N

Explanation:

Hope this helps

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Can anyone Hele please I don't know how to convert the light years in order to get an equation like the one on Alpha Centauri in

Svetradugi [14.3K]

Answer:

Log 10 ( 4.16x10^16)

Explanation:

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4 0

3 years ago

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

wlad13 [49]

Answer:

between locations that are 14 cm outboard of the chair edges
the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

(50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

67(1) +54(x) = 50(1.5)

x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

67(1.5) = 54(x) +50(1)

x = 50.5/54 ≈ 0.935 . . . . meters

<u>Scenario 1</u>

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

<u>Scenario 2</u>

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

__

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

<em>Comment on the question</em>

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

3 0

4 years ago

Using the formula d = rt, calculate the distance a car would travel at 80 mph for 2.5 hours.

hoa [83]

Using the formula,

$d= v \times t$

Here, d is distance, v is the velocity and t is time.

Given, $v = 80 \ mile /hour$ and $t = 2.5 \ hour$ .

Substituting these values in above formula, we get

$d = 80 \ mile /hour \times 2.5 \ hour = 200 \ miles$

Thus, the distance traveled by the car is 200 miles

8 0

3 years ago

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 60

Valentin [98]

Answer:

$N_T=2086285.67$

Explanation:

Given;

Thickness of the glass plate, $x=2.95\times 10^{-3}\ m$

refractive index of the glass plate, $n=1.6$

wavelength of light source in vacuum, $\lambda=600\times 10^{-9}\ m$

distance between the source and the screen, $d=1.25\ m$

Distance travelled by the light from source to screen in vacuum:

$d_v=d-x$

$d_v=1.25-0.00295$

$d_v=1.24705\ m$

So the no. of wavelengths in the vacuum:

$N=\frac{d_v}{\lambda}$

$N=\frac{1.24705}{6\times 10^{-7}}$

$N\approx2.0784\times 10^{6}$ .......................(1)

<u>Now we find the wavelength of the light wave in the glass:</u>

$n=\frac{\lambda}{\lambda'}$

where:

$\lambda'=$ wavelength of light in the medium of glass.

$1.6=\frac{600\times 10^{-9}}{\lambda'}$

$\lambda'=375\times 10^{-9}\ m=375\ nm$

Now the no. of wavelengths in the glass:

$N'=\frac{x}{\lambda'}$

$N'=\frac{2.95\times 10^{-3}}{375\times 10^{-9}}$

$N'=7.8667\times 10^{3}$ ............................(2)

From (1) & (2):

total no. of wavelengths are there between the source and the screen:

$N_T=N+N'$

$N_T=2086285.67$

5 0

3 years ago

7. A neutron at rest decays (breaks apart) into a proton and an electron. Energy is released in the decay, in the form of kineti

kondaur [170]

Answer:

The fraction of the total energy goes into kinetic energy of proton is $5.44 \times 10^{-4}$

Explanation:

Given:

Mass of proton $m_{p} = 1836 m_{e}$

Mass of electron $m_{e} = 9.1 \times 10^{-31}$ kg

Here neutron at rest decays into proton and electron

$n_{1} ^{0}$ ⇄ $P_{1} ^{1} + e_{0} ^{-1} + Q$

Where $Q =$ energy released

The $Q$ value of this reaction is given by

$Q = (\frac{m_{p} +m _{e} }{m_{e} } ) K$

Where $K =$ kinetic energy of reaction

Here we need to find fraction of the total energy released goes into the kinetic energy of the proton

$\frac{K}{Q} = \frac{m_{e} }{m_{p} + m_{e} }$

$\frac{K}{Q} = (\frac{m_{e} }{1836m_{e} + m_{e} } )$

$\frac{K}{Q} = \frac{1}{1837}$

$\frac{K}{Q} = 5.44 \times 10^{-4}$

Therefore, the fraction of the total energy goes into kinetic energy of proton is $5.44 \times 10^{-4}$

6 0

3 years ago