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brilliants [131]
3 years ago
9

1.)Which of the following would be considered an ecosystem?

Physics
2 answers:
Olenka [21]3 years ago
7 0
The answer is "a desert of rocks, sand, hawks, snakes, and scorpions" so b
Masteriza [31]3 years ago
6 0
The answer to this question is B.) because it is a variety of biotic and abiotic factors interacting with each other.
You might be interested in
A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i
laiz [17]

(a) Let <em>x</em> be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work <em>W</em> done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to <em>x</em> is

<em>W</em> = - (1/2 <em>kx</em> ² - 1/2 <em>k</em> (0.0400 m)²)

(note that <em>x</em> > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to <em>x</em>, so

<em>W</em> = ∆<em>K</em> = 0 - 1/2 <em>m</em> (0.835 m/s)²

Solve for <em>x</em> :

- (1/2 (160 N/m) <em>x</em> ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   <em>x</em> ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

<em>W</em> = - 1/2 <em>k</em> (0.0400 m)²

If <em>v</em> is the glider's maximum speed, then by the work-energy theorem,

<em>W</em> = ∆<em>K</em> = 1/2 <em>m</em> (0.835 m/s)² - 1/2 <em>mv</em> ²

Solve for <em>v</em> :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) <em>v</em> ²

==>   <em>v</em> ≈ 1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(<em>k</em>/<em>m</em>) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0
2 years ago
A car starts out traveling at 35 m/s. The car hits the brakes and decelerates at a rate of 3 m/s^2 for 5 seconds. What Distance
Ipatiy [6.2K]
Answer:

Time needed: 2.5 s
Distance covered: 31.3 m

Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

v2f=v2i−2⋅a⋅d

Isolate d on one side of the equation and solve by plugging your values

d=v2i−v2f2a

d=(15.02−10.02)m2s−22⋅2.0ms−2

d=31.3 m

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

vf=vi−a⋅t, which will get you

t=vi−vfa

t=(15.0−10.0)ms2.0ms2=2.5 s

6 0
3 years ago
A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?
madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy =\frac{1}{2} mv^2, where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0
3 years ago
Please help :)<br> i would rlly appreciate it :)))
algol [13]

Answer:

the answer for that is "A"

6 0
3 years ago
A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic
miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by v_{max}=A\omega where A is amplitude and \omega is angular velocity

Angular velocity is given by \omega=\sqrt{\frac{k}{m}}  where k is spring constant and m is mass

So v_{max}=A\sqrt{\frac{k}{m}}

A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m

3 0
3 years ago
Read 2 more answers
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