Why is it difficult to lift out the bucket form well

A 80.0-kg airplane pilot pulls out of a dive by following, at a constant speed of 180km/hr, the arc of a circle whose radius is

wolverine [178]

Answer: 8.33m/s²

Explanation:

Mass of the airplane pilot = 80kg

The speed of plane v = 180km/h = 50m/s

The radius of circle r = 300 m

The acceleration of the plane will be calculated as:

a = v²/r

a = 50²/300

a = 2500/300

a = 8.33m/s²

Note that 180km/h was converted to m/s by

= (180 × 1000)/(60 × 60)

= 180000/3600

= 50m/s

1000 meters = 1 kilometer

60 minutes = 1 hour

60 seconds = 1 minute

3600 seconds = 1 hour

6 0

3 years ago

A Hooke's law bowstring is stretched x meters until a force of f newtons is applied, and then held. By what factor will the elas

Liono4ka [1.6K]

The elastic potential energy increases by a factor of 9

Explanation:

The elastic potential energy of a bowstring is given by

$E=\frac{1}{2}kx^2$ (1)

where

k is the spring constant

x is the elongation of the bowstring

Hooke's law states the relationship between the force applied and the elongation of an elastic object:

$F=kx$

where

F is the force applied

x is the elongation

We can rewrite it as

$x=\frac{F}{k}$

And substituting into (1),

$E=\frac{1}{2}k(\frac{F}{k})^2=\frac{F^2}{2k}$

In this problem, the force applied to the bowstring is tripled,

F' = 3F

So the final elastic potential energy is:

$E'=\frac{(3F)^2}{2k}=9(\frac{F^2}{2k})=9E$

So, the elastic potential energy increases by a factor of 9.

Learn more about potential energy:

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7 0

3 years ago

An incident ray strikes a piece of diamond at an angle of 40.5 degrees. The index of refraction of air is 1.0003 and the index o

horrorfan [7]

The angle of refraction is 15.6 deg.

When a ray of light passes from medium 1 to medium 2, the refractive index of medium 2 with respect to 1 is the ratio of the absolute refractive index of medium 2 and medium 1

$_1n_2 =\frac{n_2}{n_1}$

When light passes from air into diamond, with the given values of refractive indices,

$_an_d =\frac{n_d}{n_a} =\frac{2.42}{1.0003} =2.4193$

According to Snell's law,

$_an_d=\frac{sini}{sinr} \\ sinr=\frac{sini}{_an_d} =\frac{sin(40.5)}{2.4193}=0.2684\\ r=arcsin(0.2684)=15.57 deg$

The angle of refraction is 15.6 deg

7 0

4 years ago

Read 2 more answers

object carries a charge of -8.1 ÂµC, while another carries a charge of -2.0 ÂµC. How many electrons must be transferred from the

LenKa [72]

Number of electrons transferred: $1.91\cdot 10^{13}$

Explanation:

The charge on the first object is

$Q_1 = -8.1\mu C$

while the charge on the 2nd object is

$Q_2=-2.0 \mu C$

When they are in contact, the final charge on each object will be

$Q=\frac{Q_1+Q_2}{2}=\frac{-8.1+(-2.0)}{2}=-5.05 \mu C$

So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

$\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C$

The charge of one electron is

$e=-1.6\cdot 10^{-19}C$

Therefore, the number of electrons transferred is

$N=\frac{Q}{e}=\frac{-3.05\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=1.91\cdot 10^{13}$

Learn more about electrons:

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8 0

3 years ago

Which option below correctly compares the average annual dose of background radiation to the dose liked to an increased cancer r

julia-pushkina [17]

The answer for this question would be choice "<span>B. The average annual dose of background radiation is 250 times smaller than the dose linked to increased cancer risk."

You only have to compare 4.0 x 10^-4 and 1.0 x 10^-1. And if you can observe carefully, when you try to multiply the average annual dose of background radiation by 250, you would get 0.1 which is equivalent to the amount of annual dose linked to increased cancer risk. Therefore, the answer is B.</span>

6 0

3 years ago