Answer:
False
Explanation:
Answer:
False
Explanation:
A P E X
Answer:
2.05 x 10^8 m /s
Explanation:
c = 3 x 10^8 m/s
μ = c / v
where, μ is the refractive index, c be the velocity of light in air and v be the velocity of light in the medium.
μ = 1.461
1.461 = 3 x 10^8 / v
v = 3 x 10^8 / 1.461
v = 2.05 x 10^8 m /s
The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.
To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.
First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.
Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.
So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.
#SPJ3
b. down to the floor/ground.
c. to your left.
d. to your right.
Answer:
b. down to the floor/ground
Explanation:
The direction of the angular velocity can be easily found by the use of right hand rule. This rule states that when we curl the fingers of right hand in the direction of the rotation of the object, the direction of the thumb gives the direction of angular velocity. We apply this rule to our situation. Since, the door opens outward and the hinges are on right, it means that the rotation of door is clockwise. So, when we curl the fingers of right hand in clock wise direction, the thumb will point in downward direction. So, the direction of angular velocity will be down to the floor/ground.
The correct option is:
b. down to the floor/ground
Answer:
with the current: 19.1 m/s eastern
against the current: 13.7 m/s western
Explanation:
The boat speed relative to ground is the sum of the boat speed relative to water + the water speed relative to ground.
Suppose eastern is positive, water speed due east is 2.7m/s.
When the boat is heading east, its speed relative to water is 16.4m/s, its speed relative to ground is 16.4 + 2.7 = 19.1 m/s eastern
When the boat is heading west, its speed relative to water is -16.4m/s, its speed relative to ground is -16.4 + 2.7 = -13.7 m/s western
B. fresnel
C. far-field
D. single slit
Answer:
1) I_ pendulum = 2.3159 kg m², 2) I_pendulum = 24.683 kg m²
Explanation:
In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum
Sphere
They indicate the density of the sphere roh = 37800 kg / m³ and its radius
r = 5 cm = 0.05 m
we use the definition of density
ρ = M / V
M = ρ V
the volume of a sphere is
V = 4/3 π r³
we substitute
M = ρ 4/3 π r³
we calculate
M = 37800 4/3 π 0.05³
M = 19,792 kg
Bar
the density is ρ = 32800 kg / m³ and its dimensions are 1 m,
0.8 cm = 0.0008 m and 4cm = 0.04 m
The volume of the bar is
V = l w h
m = ρ l w h
we calculate
m = 32800 (1 0.008 0.04)
m = 10.496 kg
Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is
r_cm = 1 / M (M r_sphere + m r_bar)
M = 19,792 + 10,496 = 30,288 kg
r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))
r_cm = 1 / 30,288 (5,248 + 20,7816)
r_cm = 0.859 m
This is the center of mass of the pendulum.
1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:
Sphere I = 2/5 M r2
Bar I = 1/12 m L2
parallel axes theorem
I = I_cm + m D²
where m is the mass of the body and D is the distance from the body to the axis of rotation
Sphere
m = 19,792 ka
the distance D is
D = 1.05 -0.85
D = 0.2 m
we calculate
I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168
I_sphere = 0.811472 kg m²
Bar
m = 10.496 kg
distance D
D = 0.85 - 0.5
D = 0.35 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576
I_bar = 1.5044 kg m²
The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts
I_pendulum = I_sphere + I_bar
I_pendulum = 0.811472 +1.5044
I_ pendulum = 2.3159 kg m²
this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m
2) The moment is requested with respect to the pivot point at r = 0 m
Sphere
D = 1.05 m
I_sphere = 2/5 M r2 + M D2
I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82
I_sphere = 21.84 kg m²
Bar
D = 0.5 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624
I_bar = 2,84266 kg m 2
The pendulum moment of inertia is
I_pendulum = 21.84 +2.843
I_pendulum = 24.683 kg m²
This moment of inertia is about the turning point at r = 0 m