1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:
where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is
Therefore, the angular velocity of the telescope is
2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation
where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is
Therefore, the linear velocity of the telescope is:
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.
First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055
Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3
Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.
T then is equal to 20.20 N
Answer:
Time, t = 80 seconds
Explanation:
Given that,
The frequency of the oscillating mass, f = 1.25 Hz
Number of oscillations, n = 100
We need to find the time in which it makes 100 oscillations. We know that the frequency of an object is number of oscillations per unit time. It is given by :
t = 80 seconds
So, it will make 100 oscillations in 80 seconds. Hence, this is the required solution.
Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity 
will be expressed by the following equation:
Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is:
C. The slope of the ant's displacement vs. time graph.
Answer:
V = 20.5 m/s
Explanation:
Given,
The mass of the cart, m = 6 Kg
The initial speed of the cart, u = 4 m/s
The acceleration of the cart, a = 0.5 m/s²
The time interval of the cart, t = 30 s
The final velocity of the cart is given by the first equation of motion
v = u + at
= 4 + (0.5 x 30)
= 19 m/s
Hence the final velocity of cart at 30 seconds is, v = 19 m/s
The speed of the cart at the end of 3 seconds
V = 19 + (0.5 x 3)
= 20.5 m/s
Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s
