Take 10m/s^2 for the gravitational acceleration, as we know this is a free fall, we can use the equation: d=1/2*g*t^2
Substitute g=10m/s^2, t=5s, d=125m
Answer:
Big Dipper, Little Dipper, Cassiopeia, Cepheus
Explanation:
In this region, the above constellations are circumpolar. This means that they appear above the horizon at all times. These are only visible all rear round for people living in Canada and Northern United States.
Circumpolar constellations are constellations that never appear below the horizon when seen from a particular location on planet Earth. Furthermore, these constellations can be seen all year while others are only seen at specific times during the year; thus they are known as seasonal constellations.
Five northern constellations are visible from most locations that are north of the equator. These are Cassiopeia, Cepheus, Draco, Ursa Major, and Ursa Minor.
Note that Ursa Major is often confused with the Big Dipper. While the Little Dipper (which is much fainter) is found in the Ursa Minor constellation. Cassiopeia can be recognized due to its W shape which is quite prominent.
Answer:2.7m/s^2
Explanation:
mass=2.3kg
Force=6.2Newton
Acceleration=force ➗ mass
Acceleration=6.2 ➗ 2.3
Acceleration=2.7m/s^2
Explanation:
Acceleration. Angular acceleration: Is the rate of change of the angular velocity of a body with respect to time.
Force. Torque: Is also called rotational force, since an applied torque will change the rotational motion of a body.
Mass. Moment of inertia: It is the resistance that opposes a body to rotates.
Work. Work: In a rotational motion, the work is done by the torque.
Translational kinetic energy. Rotational kinetic energy: is the kinetic energy due to the rotational motion of a body.
Linear momentum. Angular momentum: Represents the quantity of rotational motion of a body.
Impulse. Angular impulse: Is the change in angular momentum of a body.
Answer:
E = 2.87 × 10⁶ N/C
Explanation:
Electric Field Intensity Formula E = K q / r²
We will first find the charge density according the given data. (After that we will find charge enclosed by r=12 cm)
r₁ = 10 cm = 0.1 m, r₂ = 15 cm = 0.15 m, Q = 15 μC =15 × 10⁻⁶C
charge density = Charge enclosed / volume
Volume shell = 4/3 π (r₂³ - r₁³) = 4/3 π ( 0.15³ - 0.10³) =9.948 × 10⁻³ m³
charge density = 15 × 10⁻⁶C / 9.948 × 10⁻³ m³ = 1.51 × 10⁻³ C /m³
Now the Volume of shell at (r=12 cm = .12 m) = 4/3 π ( 0.12³ - 0.10³) = 3.049 × 10⁻³ m³
Now the amount of charge inside r=0.12 m which is
Charge = Charge density × volume = 0.151 × 10⁻³ C /m³ × 3.049 × 10⁻³ m³
=4.60 × 10⁻⁶ C
So E = 8.99 × 10 ⁹ N m²/C × 4.60 × 10⁻⁶ C / (0.12 m)² = 2.87 × 10⁶ N/C