Answer:
45 J
Explanation:
Assuming the level at which the ball is thrown upwards is the ground level,
We can use the equations of motion to obtain the maximum height covered by the ball and then calculate the potential energy
u = initial velocity of the ball = 3 m/s
h = y = vertical distance covered by the ball = ?
v = final velocity of the ball at the maximum height = 0 m/s
g = acceleration due to gravity = -9.8 m/s²
v² = u² + 2ay
0 = 3² + 2(-9.8)(y)
19.6y = 9
y = (9/19.6)
y = 0.459 m
The potential energy the ball will have at the top of its motion = mgh
mgh = (10)(9.8)(0.459) = 45 J
Hope this Helps!!!
Generally, the internal resistance of the new battery is small, about 0.2 euros, while the old battery is large, close to 1 euro ，
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Answer:
It depends on the size of the star but I think it's B
Explanation:
If the star is massive, it will eventually explode (supernova) and if it is a star with a high mass, the core of it will form a neutron star and if it is very massive the core will turn into a blackhole
The concept required to solve this problem is hydrostatic pressure. From the theory and assuming that the density of water on that planet is equal to that of the earth 
we can mathematically define the pressure as
Where,
= Density
h = Height
g = Gravitational acceleration
Rearranging the equation based on gravity
The mathematical problem gives us values such as:
Replacing we have,
Therefore the gravitational acceleration on the planet's surface is 
