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3 years ago

A bird of mass 1.5 kg is flying at 8 m/s at a height of 50 m above the ground. What is its

Physics

1 answer:

Bas_tet [7]3 years ago

6 0

Answer:

hyalugel puchi puchi puchi tu hello hello hello hello hello hello hello hello hello hello

Explanation:

hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello

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What does "doing work" mean when it comes to physics?

kicyunya [14]

I think it means doing work in physics

4 0

3 years ago

Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2

lbvjy [14]

-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

   = <span> 3,600 meters .</span>
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

     Distance = (1/2) · (acceleration) · (time)²

   = (1/2) · (8 m/s²)    · (30 sec)²

   = (4 m/s²) · (900 sec²)

   =     3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s². That's about 0.82 G !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.

5 0

3 years ago

A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a

Neko [114]

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

$E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$v_i$ and $v_f$ are the initial and final velocities.

With <em>m</em> = 0.175 kg,

$E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}$

The negative sign appears because energy is lost.

7 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago

Tell me about Orion nebula

RSB [31]

Is the most intensely studied celestial feature. It has also help revealed much about the process of how stars and planetary systems are formed from collapsing clouds of gas and dust. It is also the closest large star-forming region to Earth. The Orion Nebula is an enormous cloud of gas and dust, is located in our galaxy (Milky Way).

8 0

3 years ago