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3 years ago

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One of the smallest planes ever flown was the Bumble Bee II, which had a mass of 180 kg. If the pilot’s mass was 70 kg, what was

the velocity of both plane and pilot if their momentum was 20,800 kg∙m/s to the west?

1 answer:

Rom4ik [11]3 years ago

5 0

Answer:

no one help you this is bad app

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What if water didnt have a negative charge?

ArbitrLikvidat [17]

The water molecule, as a whole, has 10 protons and 10 electrons, so it is neutral. ... The unequal sharing of electrons gives the water molecule a slight negative charge near its oxygen atom and a slight positive charge near its hydrogen atoms.

8 0

3 years ago

BigorU [14]

Time = 13.5 / 2.5 = 5.4 seconds

4 0

3 years ago

Read 2 more answers

For a bronze alloy, the stress at which plastic deformation begins is 294 MPa and the modulus of elasticity is 121 GPa. (a) What

FromTheMoon [43]

Answer:

a) load in Newton is 96,138 b) 129.314mm

Explanation:

Stress = force/ area (cross sectional area of the bronze)

Force(load) = 294*10^6*327*10^-6 = 96138N

b) modulus e = stress/ strain

Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3

Strain = change in length/ original length = DL/ 129

Change in length DL = 129 * 2.34*10^ -3 = 0.31347

Maximum length = change in length + original length = 129.314mm

7 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

a 10kg ball is thrown into the air. it is going 3m/s when thrown. How much potential energy will it have at the top?

Alexandra [31]

Answer:

45 J

Explanation:

Assuming the level at which the ball is thrown upwards is the ground level,

We can use the equations of motion to obtain the maximum height covered by the ball and then calculate the potential energy

u = initial velocity of the ball = 3 m/s

h = y = vertical distance covered by the ball = ?

v = final velocity of the ball at the maximum height = 0 m/s

g = acceleration due to gravity = -9.8 m/s²

v² = u² + 2ay

0 = 3² + 2(-9.8)(y)

19.6y = 9

y = (9/19.6)

y = 0.459 m

The potential energy the ball will have at the top of its motion = mgh

mgh = (10)(9.8)(0.459) = 45 J

Hope this Helps!!!

7 0

3 years ago