Which of the following represent an atom from period 3?WILL MARK BRAINLIEST

A roller coaster car is going over the top of a 15-m-radius circular rise. The passenger in the roller coaster has a true weight

konstantin123 [22]

Answer:

v = 7.67 m/s

Explanation:

The equation for apparent weight in the situation of weightlessness is given as:

Apparent Weight = m(g - a)

where,

Apparent Weight = 360 N

m = mass passenger = 61.2 kg

a = acceleration of roller coaster

g = acceleration due to gravity = 9.8 m/s²

Therefore,

360 N = (61.2 kg)(9.8 m/s² - a)

9.8 m/s² - a = 360 N/61.2 kg

a = 9.8 m/s² - 5.88 m/s²

a = 3.92 m/s²

Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:

a = v²/r

where,

a = centripetal acceleration = 3.92 m/s²

v = speed of roller coaster = ?

r = radius of circular rise = 15 m

Therefore,

3.92 m/s² = v²/15 m

v² = (3.92 m.s²)(15 m)

v = √(58.8 m²/s²)

7 0

3 years ago

The slower the particles in a substance move,

swat32

Answer: particles move closer together

Explanation: If the motion of particles slows the particles move closer together. This is because the attraction between them pulls them toward each other. Strong attractive forces hold particles close together. As the motion of particles increases, particles move further apart.

3 0

2 years ago

What is gluteus maximum

Orlov [11]

Any of three muscles in each buttock that move the thigh, the largest of which is the gluteus maximus.

5 0

2 years ago

Read 2 more answers

Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a

lilavasa [31]

Answer:

$\lambda = 570\ nm$

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

$tan \theta = \dfrac{y}{L}$

$tan \theta = \dfrac{4}{4\times 10^3}$

$\theta = 0.0572$

Now.

$W sin \theta = m \lambda$

m = 1

$5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda$

$\lambda = 569.99 \times 10^{-9}\ m$

$\lambda = 570\ nm$

4 0

3 years ago

From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft

zhannawk [14.2K]

Answer:

$W=\frac{773}{4.45}=173.76 l b f$

Explanation:

$W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}$

The law of gravitation

$G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)$

Universal gravitational constant [S.I. units]

$m_{e}=5.976\left(10^{24}\right) k g$

Mass of Earth [S.I. units]

$m=89 kg$

Mass of a man in a spacecraft [S.I. units]

$R=6371 \mathrm{~km}$

Earth radius [km]

Distance between man and the earth's surface

$h=261 \mathrm{~km} \quad[\mathrm{~km}]$

ESULT $W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}$

$W=\frac{773}{4.45}=173.76 l b f$

4 0

2 years ago