Most of the stars in the milky way will end their lives as

10.

nlexa [21]

Answer:

Cell Membrane

Explanation:

The cell membrane controls what goes in and out of a cell, and keeps it shape, much like a city limit.

3 0

3 years ago

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. it suddenly collides directly with a stationary seal of m

Sphinxa [80]

M = mass of the whale = 1000 kg

m = mass of the seal = 200 kg

V = initial velocity of whale before collision with the seal = 6.0 m/s

v = initial velocity of the seal before collision with the whale = 0 m/s

V' = final velocity of two sea creatures after collision = ?

Using conservation of momentum

M V + m v = (M + m) V'

inserting the above values in the equation

(1000 kg) (6.0 m/s) + (200 kg) (0 m/s ) = (1000 kg + 200 kg) V'

6000 kgm/s + 0 kgm/s = (1200 kg) V'

V' = (6000 kgm/s ) /(1200 kg)

V' = 5 m/s

3 0

3 years ago

Read 2 more answers

Un movil avanza a 20 m/s y recorre una distancia de 800 km. Determinar el tiempo en horas que utiliza

Nataly_w [17]

Answer:

t = 11.1 hours

Explanation:

The question says that, "A mobile advances at 20 m / s and travels a distance of 800 km. Determine the time in hours you use".

Given that,

Speed of a mobile, v = 20 m/s = 72 km/h

Distance, d = 800 km

We know that,

Speed = distance/time

So,

$t=\dfrac{d}{v}\\\\t=\dfrac{800}{72}\\\\t=11.1\ h$

So, it will take 11.1 hours.

7 0

3 years ago

A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.

Zolol [24]

These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is $45^{\circ}$ .

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
$MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55$

Finally, the efficiency $\epsilon$ of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

3 0

3 years ago

Una bola de acero rueda y cae por el borde de una mesa desde 4ft por encima del piso. Si golpea el suelo a 5ft de la base de la

ycow [4]

Answer:

$\large\boxed{\large\boxed{10ft/s}}$

Explanation:

The question, translated, is:

A steel ball rolls and falls off the edge of a table from 4ft above the floor. If you hit the ground 5ft from the base of the table, what was your initial horizontal velocity?

<h2>Solution</h2>

This is a projectile motion, for which, the equations that you will need are:

$V_x_0= V_x=x\cdot t$

$y=y_0+Vy_o\cdot t-g\cdot t^2/2$

1. Calculate the time that it takes the ball to fall 4ft

$0=4ft-g\cdot t^2/2\\\\t^2=2\times 4ft/( 32.174ft/s^2)=0.24865s^2\\\\t=0.4986s$

2. Calculate the horizontal velocity:

$V_x_0= V_x=x\cdot t\\\\V_x_0=5ft/0.4986s=10.027ft/s\approx 10ft/s$

3 0

4 years ago