Answer:
20.1 m
Explanation:
First, find the time it takes for the cannonball to travel the horizontal distance of 50.0 m.
Given (in the x direction):
Δx = 50.0 m
v₀ = 68 cos 25° m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
(50.0 m) = (68 cos 25° m/s) t + ½ (0 m/s²) t²
t = 0.811 s
Now find the vertical displacement after that time.
Given (in the y direction):
v₀ = 68 sin 25° m/s
a = -9.8 m/s²
t = 0.811 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (68 sin 25°) (0.811 s) + ½ (-9.8 m/s²) (0.811 s)²
Δy = 20.1 m
Answer:
20.6 cm
Explanation:
charge per cm = 0.14 μC
number of electrons (e) =
to get the length of tape pulled we can apply the formula below
length of tape =
therefore we need to find the magnitude of the charge of the electrons
- 1 electron = C
now that we have the magnitude of the charge, we can find the length of the tape
- length of the tape = = 20.6 cm
Answer:
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
Explanation:
Buoyancy can be defined as a force which is created by the water displaced by an object.
Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.
Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.
The buoyancy of an object is given by the formula;
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.
Answer:
(A) = 3.57 m
Explanation:
from the question we are given the following:
diameter (d) = 3.2 m
mass (m) == 42 kg
angular speed (ω) = 4.27 rad/s
from the conservation of energy
mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1
where
Inertia (I) = 0.5mr^{2}
ω = \frac{v}{r}
equation 1 now becomes
mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}
gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}
4gh = 2v^{2} + v^{2}
h = 3v^{2} ÷ 4 g .... equation 2
from ω = \frac{v}{r}
v = ωr = 4.27 x (3.2 ÷ 2)
v = 6.8 m/s
now substituting the value of v into equation 2
h = 3v^{2} ÷ 4 g
h = 3 x (6.8)^{2} ÷ (4 x 9.8)
h = 3.57 m