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4 years ago

Why is AU (astronomical unit) the unit of distance that we used instead of kilometers?

Physics

1 answer:

Novay_Z [31]4 years ago

7 0

Answer:

The solar system is enormous, and interstellar space is even bigger. One astronomical unit is equal to 150 million kilometers. This makes it much easier to count the distances if they're in counts of Astronomic Units instead of having to count everything in millions or billions of kilometers

Explanation:

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Compressing the air by squeezing the bottle was accompanied by a(n) ________ in the temperature of air inside the bottle

Georgia [21]

Increase in temprature

3 0

3 years ago

a 75-kg refrigerator is located on the 70th floor of a skyscraper (300 meters above ground). what is the potential energy of the

mart [117]

Here, you can calculate it's potential energy with respect to ground.
We know, U = mgh
Here, m = 75 Kg
g = 9.8 m/s² [ constant value for earth system ]
h = 300 m

Substitute their values into the expression:
U = 75 × 9.8 × 300
U = 220500 J

In short, Your Final Answer would be 220,500 J

Hope this helps!

8 0

3 years ago

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f

IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0

3 years ago

An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os

oksian1 [2.3K]

Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

Spring Constant:

As w= $\sqrt{k/m}$

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)

⇒ v= $\sqrt{k A^2/m}$

⇒ v= $\sqrt{179.2 * 0.27/ 0.628}$ \

⇒ v= 8.78 m/s

Maximum force will be exerted on the block when it is at maximum distance.

F= k* A ( A is amplitude of oscillation)

F= 179.2 * 0.27 N

F= 48.4 N

5 0

3 years ago

Q 36 - URGENT HELP PLS<br> WILL BE MARKED AS BRAINLIEST.

Sliva [168]

The answer is B.

This is because you add up all of the times (1.44s+1.70s+1.58s+1.76s) and you get 6.48 then you divide 6.48 by 4 to get the average of the times. Now you get the distance (200m) and because speed=distance/time you divide 200m/1.62s to get 123m/s. I hope this made sense :)

5 0

3 years ago