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3 years ago

Which does not contain a lens?

Physics

2 answers:

Doss [256]3 years ago

8 0

Answer:

Explanation:

Nutka1998 [239]3 years ago

4 0

<span>A lens is a piece of glass or other transparent substance with curved sides for concentrating or dispersing light rays, used singly (as in a magnifying glass) or with other lenses (as in a telescope). A binoculars, an eye, and a camera all contains a lens. A mirror does not contain a lens.</span>

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Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago

The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl

Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, $m_1=5\ kg$

Initial speed of block A, $u_1=5\ m/s$

Mass of block B, $m_2=8\ kg$

Initial speed of block B, $u_2=0$

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s$

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0

3 years ago

If you have 12 atoms of hydrogen before a chemical reaction how many atoms of hydrogen will be president after the chemical reac

kaheart [24]

If you have 12 atoms of hydrogen before a chemical reaction, the number of hydrogen atoms that will be present after the chemical reaction is 12 atoms.
The Law of Conservation of Mass (LOCOM) states that mass is neither created nor destroyed before and after any chemical reaction.
According to the Law of Conservation of Mass (LOCOM), a balanced chemical equation requires that the number of atoms on the reactant side must be equal to the number of atoms on the product side of any chemical reaction.
In this context, a chemical reaction having 12 atoms of hydrogen as reactants at the beginning, should also produce a total of 12 atoms of hydrogen as products at the end of the chemical reaction.

4 0

2 years ago

a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an

Mademuasel [1]

We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
$T_1 - W_p - W_s + T_2 =0$
where
$W_p = (90kg)(9.81 m/s^2)=883 N$ is the weight of the person
$W_s = (200kg)(9.81 m/s^2)=1962 N$ is the weight of the scaffold
Re-arranging, we can write the equation as
$T_1 = 2845 N-T_2$ (1)

2) Equilibrium of torques:
$T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0$
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using $W_p = 883 N$ and replacing T1 with (1), we find
$2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0$
from which we find
$T_2 = 1128 N$

And then, substituting T2 into (1), we find
$T_1 = 1717 N$

8 0

3 years ago

Use these words in a sentence proton neutron and isotope

Hunter-Best [27]

"The proton and neutron have nothing to do with the isotope little billy"

4 0

3 years ago