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maksim [4K]
3 years ago
5

Suppose that the angular separation of two stars is 0.1 arcseconds, and you photograph them with a telescope that has an angular

resolution of 1 arcsecond. How will the stars appear in the photograph?
Physics
1 answer:
Crank3 years ago
4 0

Answer: the photograph will likely show only one star.

Explanation:

Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.

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A book prone to air resistance is released from rest 300 m
yaroslaw [1]

Answer:

Approximately 73\%.

(Assuming that g = \rm 9.81\; m \cdot s^{-2}.)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let m(\text{book}) represent the mass of the book.

The initial potential energy of the book is  

\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 \times 300) \cdot m(\text{book})\cr &= \left(2.943\times 10^3\right) \cdot m(\text{book})\end{aligned}.

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be (2.943\times 10^3) \cdot m(\text{book}).

When the book was about to hit the ground, its speed is \rm 40\; m \cdot s^{-1}. Its kinetic energy would be:

\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}.

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be \left(8.00\times 10^2\right)\cdot m(\text{book}) when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}.

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, m(\text{book}), was eliminated in this process.

\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}.

5 0
3 years ago
The amount of matter I’m an object is know as
Oduvanchick [21]

The amount of matter I’m an object is know as <em>the object's mass</em> .

8 0
4 years ago
How do you calculate force?
DIA [1.3K]
Force (N) = mass (kg) × acceleration (m/s²)
6 0
4 years ago
Which of the following best describes each particle?
Nana76 [90]

Answer:

C. Protons are positive, neutrons are neutral, and electrons are negative

4 0
3 years ago
For each 1 degree Celsius that temperature incileases, the speed of<br> sound ___ by 0.6 m/s.
Levart [38]

Answer:

A: For each 1 degree Celsius that temperature decreases, the speed of sound decreases by 0.6 m/s. So sound travels through dry, -20 °C air at a speed of 319 m/s

Explanation:

6 0
3 years ago
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