6.78 x 10^8 MHz
5.36 x 10^6 kV
1500mA
Calculate the following numbers into scientific notation (include units)
7800nm + 95pm =
2500nm - 7pm =
(65x10^4m) x (4.5x10^-6s-1)=
(24x10^5m) / (2x10^-8s) =
convert the following to the prefix commonly used (use units)
55 000 000 Hz
4.8x10^4V
0.0068s
0.00000000234m
Answer:
28.9 m/s
4.52 s
Explanation:
Given in the y direction:
s = 25 m
u = U sin 50°
v = 0 m/s
a = -9.8 m/s²
Find: U
v² = u² + 2as
0² = (U sin 50°)² + 2 (-9.8) (25)
U = 28.9 m/s
Given in the y direction:
s = 0 m
u = 28.9 sin 50° = 22.1 m/s
a = -9.8 m/s²
Find: t
s = ut + ½ at²
0 = (22.1) t + ½ (-9.8) t²
t = 4.52 s
Answer:
The centripetal acceleration that the moon experiences will be almost equal to the gravitational force that the Earth does in the moon,
Now, remember these two things:
F = m*a
and Fg = G*M1*M2/r^2
the first equation says that the force applied to something is equal to the mass of the object times the acceleration.
The second equation is for the gravitational force, where G is a constant, M1 and M2 are the masses of both objects, in this case, the Earth and the moon, and r is the distance.
We know that the acceleration in the surface of the Earth is:
a = Fg/M2 = g = G*M1/(RE)^2
now, for the moon we will have:
a = G*M1/(60RE)^2 = (G*M1/(RE)^2) *(1/60^2)
Here the term in the left is equal to g, so we have:
(G*M1/(RE)^2) *(1/60^2) = g*(1/60^2)
So the centripetal acceleration of the moon is 60^2 = 3600 times smaller than g.
The object will sink, because it is more dense than water.
Let's see this in detail. There are two forces acting on the object:
- its weight, which points downward, given by
where is the object's density, is its volume, and g is the gravitational acceleration.
- The buoyancy force, which points upward, given by
where is the water density, is the volume of water displaced by the object.
We see that it is always , so the object will sink. In fact:
. We are told the object's density is 1.05 g/mL, while the water density is 1.00 g/mL.
: the two volumes are equal when the object is completely submersed, and the volume of water displaced cannot be greater than the volume of the object.
So, W > B, and the object will sink.
the load distance is approximately 0.2 meters or 20 cm. To calculate the load distance in this scenario, we can use the principle of moments, which states that the sum of clockwise moments about a point is equal to the sum of counterclockwise moments about the same point when an object is in equilibrium.
In this case, we have two forces:
1. A known force of 325 N applied at a distance of 40 cm (0.4 meters) from the point of interest.
2. An unknown force of 650 N applied at an unknown distance, which we need to find.
Let's assume the unknown distance is "d" meters.
Using the principle of moments:
Clockwise Moment = Counterclockwise Moment
(325 N) * (0.4 m) = (650 N) * (d m)
Solving for "d":
d = (325 N * 0.4 m) / 650 N
d ≈ 0.2 meters
So, the load distance is approximately 0.2 meters or 20 cm.
To know more about moments:
#SPJ1
b. False
the correct answer is true.