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3 years ago

What is a an reason why a object gets extinct

Physics

1 answer:

adoni [48]3 years ago

7 0

Answer:

An object may become extinct due to the lack of quantity of the object and the object no longer exists so it is now extinct.

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A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]

inna [77]

This is a defective, misleading question, and should never be asked in a Physics class.

There is no such thing as the force due to the impact.

If you know how long it takes the clam to stop once it begins to hit the dirt,
then you can calculate the impulse transferred to it, and tease a force out
of that. But the question doesn't give us the time.

It depends on the material of the surface. Was the clam dropped onto dirt ?
Into a dumpster ? Onto grass ? Concrete ? Styrofoam ? Mud ? The answer
is different in each case, and we still need to know the short length of time
AFTER it first encountered whatever surface brought it to rest.

I would kick this question back to the Physics teacher. It's meaningless,
and the longer you try to work on it, the more nonsense you'll plant into
your head that'll need to be dug out later.

8 0

3 years ago

A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I

melamori03 [73]

(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
$v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s$

(b) The initial speed of the skater is
$v_i = 4 m/s$
while the final speed is
$v_f = 5.3 m/s$
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
$a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2$

(c) The initial speed of the skater is
$v_i = 13.0 m/s$
while the final speed is
$v_f=0$
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
$2aS=v_f^2 -v_i^2$
from which we find
$a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2$
where the negative sign means it is a deceleration.

4 0

3 years ago

A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic

Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light $c= 2.99793\times10^{8}\ m/s$

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

$\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}$

Put the value into the formula

$\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}$

$\Delta l=1\sqrt{1-(0.96)^2}$

$\Delta l=0.28\ m$

We need to calculate the time the stick take to move

Using formula of time

$t=\dfrac{\Delta l}{v}$

Put the value into the formula

$t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}$

$t=9.72\times10^{-10}\ sec$

$t=0.97\ ns$

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0

3 years ago

A football is kicked from ground level at an angle of 53 degrees. It reaches a maximum height of 7.8 meters before returning to

Nonamiya [84]

Answer:

1.61 second

Explanation:

Angle of projection, θ = 53°

maximum height, H = 7.8 m

Let T be the time taken by the ball to travel into air. It is called time of flight.

Let u be the velocity of projection.

The formula for maximum height is given by

$H = \frac{u^{2}Sin^{2}\theta }{2g}$

By substituting the values, we get

$7.8= \frac{u^{2}Sin^{2}53 }{2\times 9.8}$

u = 9.88 m/s

Use the formula for time of flight

$T = \frac{2uSin\theta }{g}$

$T = \frac{2\times 9.88\times Sin53 }{9.8}$

T = 1.61 second

7 0

2 years ago

A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm

Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

$v =\sqrt{\dfrac{T}{\mu}}$

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

$v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}$

v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

λ=2 L

λ=2 x 0.577 = 1.154 m

we now,

v = f λ

$f = \dfrac{v}{\lambda}$

$f = \dfrac{1074.49}{1.154}$

f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0

3 years ago