The formula for orbital speed v is v=(G*Me/r)^1/2
Wher G= 6.67E-11, Me= 6E24, r= Re+h= 6.4E6+740000
putting values in the formula we get
v= 7486.7 m/s or v= 7.4867 km/s
Answer:
Controlled braking
Explanation:
CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.
Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.
Answer:
The value is 
Explanation:
From the question we are told that
The diameter of each wheel is 
The mass of the motorcycle is 
The rotational kinetic inertia is 
The mass of the rider is 
The velocity is 
Generally the radius of the wheel is mathematically represented as
=> 
=> 
Generally from the law of energy conservation
Potential energy attained by system(motorcycle and rider ) = Kinetic energy of the system + rotational kinetic energy of both wheels of the motorcycle
=> 
=> 
Here 
is the angular velocity which is mathematically represented as
So
Here 
![395 * 9.8 * h = 0.5 * 395 * (23.61)^2 + 2.1 *[\frac{ 23.61}{ 0.26} ] ^2](https://tex.z-dn.net/?f=395%20%2A%20%209.8%20%2A%20%20h%20%20%3D%20%20%200.5%20%20%20%20%2A%20%20%20395%20%2A%20%20%2823.61%29%5E2%20%2B%20%202.1%20%20%2A%5B%5Cfrac%7B%2023.61%7D%7B%200.26%7D%20%5D%20%5E2)
=>
Answer
-8.67× 10^6 N/C
Explanation:
The Electric Field is defined as force per unit charge.
E = Q/ 4π£r2
Qv= −6.5 μCm3
Qv = Q/ V= Q/ 4/3 πr3
Hence Q = 4/3 πr3 × Qv
Hence E = 4/3 πr3 × Qv / 4π£r2= Qvr/3£
−6.5 μ × 4/ 3×8.854 ×10^-12
-6.5 × 4 × 10^6/3 = -8.67× 10^6 N/C
Note: £ = 8.854×10^-12m/F
is the permittivity of free space
Qv is the charge per unit volume
V is volume and volume
Answer:
Explanation:
As we know that due to a thin sheet electric field at a point near its surface is given as
now we have two sheets of opposite charges so the net electric field is double that of field due to one sheet
so we have
now we have
