Ask question

Ask question

All categories

PSYCHO15rus [73]

3 years ago

12

A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 4.0\,\text m4.0m4, point, 0, start text, m, end

text before it lands on the dog. We can ignore air resistance. How many seconds did the acorn fall?

1 answer:

kirill [66]3 years ago

8 0

Answer:

0.903 seconds

Explanation:

To find how many seconds the acorn fall, we can use the formula for distance travelled with constant acceleration:

D = Vo*t + a*t^2/2,

where D is the distance travelled, Vo is the inicial speed, t is the time and a is the acceleration.

In our problem:

Vo = 0,

a = g = 9.81 m/s2,

D = 4 meters.

So, we can solve the equation to find the time:

4 = 0*t +9.81*t^2/2

4.905*t^2 = 4

t^2 = 4/4.905 = 0.8155

t = 0.903 seconds

You might be interested in

A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

elena-14-01-66 [18.8K]

Answer:

<u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

<u>2. Formulae:</u>

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

<u>3. Solution (calculations)</u>

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

3 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

2 years ago

The number of sound waves per unit time is called what

Andreyy89

Answer:

Frequency

Explanation:

Hoped this helped

3 0

3 years ago

Newton’s third law in football?

trapecia [35]

Newtons third law of motion applies to football as when the ball is thrown, force is applied to the football thrown forward and also force back to the player. Therefore it states that every action has a reaction. The player does not get a huge force back from throwing the football due to the player having a bigger mass than the football being thrown.

In short, when the football is throw, there is a reaction with the ball and the player accordingly with the force put into the ball.

6 0

3 years ago

What is the difference between an object’s speed and its acceleration? NO LINKS

Arisa [49]

Answer: Hello! An objects speed is constant and has the units meters per second (m/s); thus, it does not change overtime. Acceleration is a rate of change where the speed does either increase or decrease overtime from its inital value; its units are meters per second second (m/s/s). I hope that helps!

5 0

2 years ago