Ask question

Ask question

All categories

3 years ago

10

A bird flying 5.0 m overhead sees me drop it, and starts to dive straight down towards the bread the instant I release it. She c

atches it after it falls 3.0 m. Assuming she accelerates constantly from rest (0 = 0) at the time I let go of the bread, what is her acceleration?

1 answer:

gladu [14]3 years ago

5 0

Answer:

15

Explanation:

muntiply the 5.0.3.0 ,you will get ans

You might be interested in

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

In a lab, a 5 L sample of gas with a pressure of 6 atm was transferred to a one liter flask.

wolverine [178]

Answer:

30 atm

Explanation:

P1V1 = P2V2

P2 = P1V1/V2

P2 = (6 atm * 5 L) / 1L

P2= 30 atm

6 0

3 years ago

Describe the difference between renewable and nonrenewable resources. Give at least two advantages and two disadvantages of each

elixir [45]

Answer: (A), (C), and (E)

Explanation:

5 0

3 years ago

Use the drop-down menus to answer the questions.

Softa [21]

Answer:

Use the drop-down menus to answer the questions.

Which fossil is the oldest?

✔ trilobite

Which fossil is the youngest?

✔ ammonite

Explanation:

ye

6 0

3 years ago

Read 2 more answers

the pilot of a new stealth helicopter, which has a mass of 15000 kg and was traveling at 180 m/s, accelerated to 250 m/s in 6 s

nirvana33 [79]

Answer:

Force = 175000 newtons

Final momentum = 1050000 N*S

Explanation:

We know that force is equal to Mass * acceleration

We also know that the mass of object is 15,000kg.

We know acceleration is (change in velocity)/time

Our initial velocity is 180m/s

our final velocity is 250 m/s

the time it took was 6 seconds

our change in velocity is 250 - 180 = 70

the time it took was 6 seconds

a = 70/6 = 11.67 m/s^2

So we can calculate force, by doing Mass * Acceleration, which is 15,000 * 11.67

Force is 175000 Newton's

Momentum is calculated by Velocity * mass or Force * time

we could do either one

<u>velocity</u><u> </u><u>*</u><u> </u><u>mass </u>way

we are given final velocity is 250 m/s, and that the initial velocity is 180 m/s. We first have to find the change in velocity by subtracting final and initial and get 70, and then we multiply that by the mass, which is 15000 kg.

we multiply them together and get 1050000 Kg*m/s

<u>Force </u><u>*</u><u> </u><u>time </u>way

We just calculated the force above, and got 175050 newton's. from that we just multiply by the time, which is 6 seconds

and we get a result of

1050000 Newton*seconds

*** remember newton*seconds is the same as kg*m/s

8 0

3 years ago