Why would the kinetic energy at the bottom of the track be less than the potential energy at the top of the track?

Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.2cm and a standard deviation o

Svet_ta [14]

Answer:

97.5%

Explanation:

By the empirical rule (68-95-99.7),

68% of data are within μ - σ and μ + σ
95% of data are within μ - 2σ and μ + 2σ
99.7% of data are within μ - 3σ and μ + 2σ

σ and μ are the standard deviation and the mean respectively.

From the question,

μ = 7.2 cm

σ = 0.38 cm

7.96 = 7.2 + (n × 0.38)

n = 2

Hence, 7.96 represents μ + 2σ.

P(X < μ + 2σ) = P(X < μ) + P(μ < X < μ + 2σ)

P(X < μ) is the percentage less than the mean = 50%.

P(μ < X < μ + 2σ) is half of P(μ - 2σ < X < μ + 2σ) = 95% ÷ 2 = 47.5%.

Considering this, for apples that are no more than 7.96 cm,

P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%

5 0

3 years ago

While studying bacterial cells, scientists measure the lengths of the cells in one colony. The chart shows their data.

Hitman42 [59]

The scientists should best deal with this measurement by stating that there was an error during measuring and collect further data.

5 0

3 years ago

Read 2 more answers

You have two vectors, which are 2.59 m at 30.0° north of east and 4.18 m at 60.0° north of west. What is the magnitude in meters

andrew11 [14]

Answer: $\ec{r}=0.153\hat{i}+4.914\hat{j}$ v

Explanation:

Given

Vector 1

$\vec{a}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )$

Vector 2

$\vec{b}=4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )$

Resultant $\vec{r}=\vec{a}+\vec{b}$

$\vec{r}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )+4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )$

$\vec{r}=0.153\hat{i}+4.914\hat{j}$

$|r|=4.916$

5 0

3 years ago

A wire is stretched between two posts. Another wire is stretched between two posts that are four times as far apart. The tension

Elena-2011 [213]

Answer:

Therefore,

The speed of the wave on the longer wire is 95 m/s.

Explanation:

Given:

For Short wire, speed is

$v_{s}=190\ m/s$

Let length of Short and Longer wire be $L_{s}\ and\ L_{l}$ such that

$L_{l}=4\times L_{s}$

To Find:

$v_{l}=?$ Speed on the longer wire

Solution:

The speed of a pulse or wave on a string under tension can be found with the equation,

$v=\sqrt{\dfrac{F_{T}\times L}{m}$

Where,

$F_{T}$ = Tension on the wire

L = Length of Sting

m = mass of String

So here we have,

$F_{T}$ = same

$L_{l}=4\times L_{s}$

Therefore,

$v_{s}=\sqrt{\dfrac{F_{T}\times L_{s}}{m}$ ......equation ( 1 )

And

$v_{l}=\sqrt{\dfrac{F_{T}\times L_{l}}{m}$ .......equation ( 2 )

Dividing equation 1 by equation 2 and on Solving we get

$\dfrac{v_{s}}{v_{l}}=\sqrt{\dfrac{L_{s}}{L_{l}}}$

Therefore,

$v_{l}=v_{s}\sqrt{\dfrac{4\times L_{s}}{L_{s}}}=190\times 2=380\ m/s$

Therefore,

The speed of the wave on the longer wire is 95 m/s.

8 0

4 years ago

a proton travelling along is x-axis is lowed by a niform electric field E. at x = 20.0 cm, the proton has a speed of 3.5x10^6 m/

anzhelika [568]

Answer:

Magnitude of electric field is 1.06 x $10^5$ V/m along negative X-direction

Explanation:

Given: initial velocity of proton = u = 3.5 x $10^6$ m/s

final velocity of proton = v = 0 m/s

initial point $l_i$ = 0.2 m and final point is $l_f$ = 0.8 m

According to conservation of energy:

change in in kinetic energy = change in potential energy of proton

⇒ $\frac{m}{2}(v^2-u^2 ) = qE(l_i - l_f)$

where q and m is the charge and mass of proton E is the electric field , $l_i$ and $l_f$ is the initial and final position of proton

on substituting the respected values we get,

1.023 x $10^-^1^4$ = 9.6 x $10^-^2^0$ x E

⇒ E = 1.06 x $10^5$ V/m

external force is opposite to the motion as velocity of proton decreases with distance.

Therefore, magnitude of electric field is 1.06 x $10^5$ V/m along negative X-direction

5 0

3 years ago