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3 years ago

The study of the living and non living things in a environment is

Physics

1 answer:

statuscvo [17]3 years ago

4 0

Answer:

Ecosystem.

Explanation:

An ecosystem is where there are both living and nonliving things.

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Spacecrafts have landed on the surface of which planet?

ki77a [65]

I'm pretty sure its Venus!!!

7 0

3 years ago

A penny is dropped off the top of the Stratosphere from rest and falls freely with the

Roman55 [17]

Answer:

S = 122.5m

Explanation:

Given the following data;

Acceleration due to gravity = 9.8m/s²

Time, t = 5 seconds

Since it's a free fall, initial velocity, u = 0

To find the displacement, we would use the second equation of motion given by the formula;

$S = ut + \frac {1}{2}at^{2}$

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

$S = 0*5 + \frac {1}{2}*(9.8)*5^{2}$

$S = 0 + 4.9*25$

S = 122.5m.

3 0

3 years ago

Chapter 36, Problem 007 Light of wavelength 586 nm is incident on a narrow slit. The angle between the first diffraction minimum

svp [43]

Answer:

$d =4.77\times 10^{-5}\ m$

Explanation:

given

wavelength of light λ = 479 nm

= 479 x 10⁻⁹ m

the angle

θ = 1.15 / 2 = 0.575°

using

condition for diffraction minimum ,

d sinθ = m λ

for first minimum m = 1

d sinθ = λ

therefore ,

slit width

$d =\dfrac{\lambda}{sin\theta}$

$d =\dfrac{479\times 10^{-9}}{sin 0.575^0}$

$d =\dfrac{479\times 10^{-9}}{0.01}$

$d =4.77\times 10^{-5}\ m$

hence, the width of the slit is equal to $d =4.77\times 10^{-5}\ m$

3 0

2 years ago

Two examples of a stopping motion

Anton [14]

Pause?Freeze?
Stop?
Halt?

5 0

2 years ago

Read 2 more answers

A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,

Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m)

= 0.15 √(300/1)

= 2.598 m/s.

b) Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.

This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0

3 years ago