List out any 4 objectives of education

An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

SCORPION-xisa [38]

Answer:

$a_y=0.92m/s^2$

Explanation:

To solve this problem we use the formula for accelerated motion:

$y=y_0+v_{y0}t+\frac{a_yt^2}{2}$

We will take the initial position as our reference ( $y_0=0m$ ) and the downward direction as positive. Since the rock departs from rest we have:

$y=\frac{a_yt^2}{2}$

Which means our acceleration would be:

$a_y=\frac{2y}{t^2}$

Using our values:

$a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2$

5 0

3 years ago

Which of the following activities could lead to injuries?

Ipatiy [6.2K]

Answer:

Fighting

Explanation:

Fighting with someone like punch for punch can make the person bleed which an injury to the guy who you fought with and you can be arrested if the parents file a complain at the police station

6 0

3 years ago

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A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of

mina [271]

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m; &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

= (2.1 × 10⁷ + 6.37 × 10⁶) m

= 27370000

= 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )

= 3818.215

= 3.818 × 10³

= 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × $\frac{r}{V_o}$

= 2 × 3.14 × $\frac{2.737*10^7}{3.818*10^3}$

= 45019.28

= 4.5 × 10 ⁴ s

6 0

3 years ago

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o

Ivahew [28]

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

(c) $\mu=9.71*10^{-24}A\cdot m^2$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$