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3 years ago

What is the shortest distance between two points

Physics

1 answer:

White raven [17]3 years ago

7 0

Answer: displacement

Explanation:

According to the definition of displacement it is the shortest distance between two points.

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If the focal length of a concave mirror is 18cm, find its radius of curvature.

Galina-37 [17]

Given :

The focal length of a concave mirror is 18 cm.

To Find :

The radius of curvature of the concave mirror.

Solution :

We know,

$\text{Focal length}=\dfrac{\text{Radius of curvature}}{2}\\\\F=\dfrac{R}{2}\\\\R = 18\times 2\ cm\\\\R = 36 \ cm$

Therefore, the radius of curvature of concave mirror is 36 cm.

Hence, this is the required solution.

8 0

3 years ago

Select the correct answer.

Simora [160]

Answer:

A. 2.36 Newtons

Explanation:

F = GmM/d²

F = 6.673 x 10⁻¹¹(1)(5.98 x 10²⁴) / (1.3 x 10⁷)²

F = 2.36121...

Very poor question design.

mass of box... 1 significant digit

distance... 2 significant digits

mass of earth... 3 significant digits

value of G... 4 significant digits

Answer precision to 3 significant digits is not justifiable

7 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant