All categories

3 years ago

According to astronomers, how many galaxies are distributed across the observable universe?

Physics

2 answers:

Allushta [10]3 years ago

7 0

Answer:actually it's 120 billion so you can choose 100 billion

Explanation:

Kay [80]3 years ago

4 0

B: 1 trillion distributed across the universe. If I remember correctly.

You might be interested in

PLEASE HELP!! WILL REWARD!!

Fudgin [204]

$velocity = \frac{displacement}{time}$

Displacement measures how far it is from the starting point and the final location, which in this case is the distance between Honolulu and Los Angeles. The difference between distance and displacement is that distance measures how far the object actually travels. For example, the airplane may not fly in a straight line and it may have detours, making the distance much greater than the displacement.

On the other hand, the plane could fly at different speeds at different periods of time. It will be so much more complicated if we have to calculate it. That's why this question only asks you to calculate the average velocity of the plane, ie. $\frac{total displacement}{total time}$

So, we have

the required average velocity = $\frac{2558}{5}$

= 511.6 miles/hour

Hope this helps!

6 0

3 years ago

Read 2 more answers

A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m>s at an angle of 33.0%1b above th

Brilliant_brown [7]

The motion described here is a projectile motion which is characterized by an arc-shaped direction of motion. There are already derived equations for this type of motions as listed:

Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)

where

Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height

Part A.

Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m

Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:

y = v₀t + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds

a = (v₁-v₀)/t
9.81 = (v₁ - 30)/0.839
v₁ = 38.23 m/s

Part C.
y = xtanθ + gx²/(2v₀²cos²θ)
Hmax + 15 = xtanθ + gx²/(2v₀²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m

3 0

3 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

6 0

3 years ago

An increase in the entropy of a system can be described as an increase in the total amount of _____ of a system.

fredd [130]

An increase in the entropy of a system can be described as an increase in the total amount of disorder in a system.

Entropy

Entropy is a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty. From classical thermodynamics, where it was originally recognized, through the microscopic description of nature in statistical physics, to the fundamentals of information theory, the phrase and concept are utilized in a variety of disciplines. It has numerous applications in physics and chemistry, biological systems and how they relate to life, cosmology, economics, sociology, weather science, and information systems, especially the exchange of information.

To learn more about entropy refer here:

brainly.com/question/13146879

#SPJ 4

6 0

1 year ago

Two vectors have magnitudes 20 m and 44 m. Which of the following cannot possibly be the magnitude of the resultant of the two v

Volgvan

Answer:

44M 64M

Explanation:

4 0

3 years ago