Mercury is in the 80th position in the periodic table. How many protons does it have?

HELP ME PLEASE I NEED IT NOW

tia_tia [17]

Answer:

the angle of incident is 40°

Explanation:

NQ is the normal to the mirror, therefore

angle NQA =90°

PQA = 50°

incident angle = NQA - PQA

90°- 50° = 40°

note that the angle of reflection is equal to the angle of incident

8 0

2 years ago

The benefits of jumping rope include: (which of these things)

Svetach [21]

Answer:

E. All of the Above

Explanation:

By doing any kind of exercise or physical activity, you are increasing your overall health. Your muscle strengthen because of the jumping and movement of arms. You are more alert because you have to time each jump right in order to keep going. By breathing evenly while jumping, you do help your Cardiorespiratory fitness as well. And of course, you increase your athletic ablility over all with much endurance and practice.

3 0

2 years ago

Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth

mylen [45]

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

4 0

2 years ago

Read 2 more answers

What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m

Pepsi [2]

v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

5 0

3 years ago

Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the

Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1 ⇒ w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

L₀ = L_f

I₀ w₀ = I w

w = $\frac{I_o}{I}$ w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

I <I₀

I₀/I > 1 ⇒ w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0

2 years ago