Answer: 5,640 s (94 minutes)
Explanation:
the tangential speed of the HST is given by
(1)
where
is the length of the orbit
r is the radius of the orbit
T is the orbital period
In our problem, we know the tangential speed:
. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:
![r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m](https://tex.z-dn.net/?f=r%3D6.38%5Ccdot%2010%5E6%20m%2B569%2C000%20m%3D6.95%5Ccdot%2010%5E6%20m)
So, we can re-arrange equation (1) to find the orbital period:
![T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B2%20%5Cpi%20r%7D%7Bv%7D%3D%5Cfrac%7B2%20%5Cpi%20%286.95%5Ccdot%2010%5E6%20m%2Fs%29%7D%7B7%2C750%20m%2Fs%7D%3D5%2C640%20s)
Dividing by 60, we get that this time corresponds to 94 minutes.
Answer:
![\mu_s = 0.62](https://tex.z-dn.net/?f=%5Cmu_s%20%3D%200.62)
![\mu_k = 0.415](https://tex.z-dn.net/?f=%5Cmu_k%20%3D%200.415)
The motion of the block is downwards with acceleration 1.7 m/s^2.
Explanation:
First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.
![x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2](https://tex.z-dn.net/?f=x%20-%20x_0%20%3D%20v_0t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C3.4%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7Da%282%29%5E2%5C%5Ca%20%3D%201.7~m%2Fs%5E2)
Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.
![F = ma\\F = 1.7m](https://tex.z-dn.net/?f=F%20%3D%20ma%5C%5CF%20%3D%201.7m)
Now, let’s investigate the free-body diagram of the block.
Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,
![F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415](https://tex.z-dn.net/?f=F%20%3D%20mg%5Csin%28%5Ctheta%29%20-%20%5Cmu_k%20mg%5Ccos%28%5Ctheta%29%5C%5C1.7m%20%3D%20mg%5Csin%2831.8%29%20-%20%5Cmu_k%20mg%5Ccos%2831.8%29%5C%5C1.7%20%3D%20%289.8%29%5Csin%28%5Ctheta%29%20-%20mu_k%289.8%29%5Ccos%28%5Ctheta%29%5C%5Cmu_k%20%3D%200.415)
As for the static friction, we will consider the angle 31.8, but just before the block starts the move.
![mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62](https://tex.z-dn.net/?f=mg%5Csin%2831.8%29%20%3D%20%5Cmu_s%20mg%5Ccos%2831.8%29%5C%5C%5Cmu_s%20%3D%20tan%2831.8%29%20%3D%200.62)
The rubber protects him from being electrocuted by the flow of current going through the plug.
Hope this helped!!
The acceleration of the bus is 1.11 meters per second square to the direction of motion
Explanation:
Acceleration is the rate of change of velocity
The formula of the acceleration is
, where
is the initial velocity
is the final velocity- t is the time
A bus that goes from 10 km/h to a speed of 50 km/h in 10 seconds
→
= 10 km/h
→
= 50 km/h
→ t = 10 seconds
Change the unite of the time from seconds to hour
→ 1 hour = 60 × 60 = 3600 seconds
→ 10 seconds =
hour
Substitute these values in the formula of the acceleration above
→ ![a=\frac{50-10}{\frac{1}{360}}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B50-10%7D%7B%5Cfrac%7B1%7D%7B360%7D%7D)
→ a = 14400 km/h²
To change the unit of acceleration to meter per second change the
kilometer to meter and the hour to seconds
→ 1 km = 1000 m
→ 1 hour = 3600 seconds
→ ![a=\frac{14400*1000}{(3600)^{2}}=\frac{10}{9}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B14400%2A1000%7D%7B%283600%29%5E%7B2%7D%7D%3D%5Cfrac%7B10%7D%7B9%7D)
→ a = 1.11 m/sec².
The acceleration of the bus is 1.11 meters per second square to the direction of motion
Learn more:
You can learn more about the acceleration in brainly.com/question/6323625
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