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Answer:
3/7 ω
Explanation:
Initial momentum = final momentum
I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet
-Iω + 6Iω - 2Iω = 7I ωnet
3Iω = 7I ωnet
ωnet = 3/7 ω
The final angular velocity will be 3/7 ω counterclockwise.
Answer:
Use a faster than normal approach and landing speed.
Explanation
For pilots, it is one of the critical moments of the flight that concentrates 12% of fatal accidents. The main difficulty lies in reaching enough speed to take flight within the space of the runway. At present, it ceased to be a challenge for the aircraft, since the engine power improved, so the takeoff ceased to be the most dangerous moment of the flight.
One of the risks that aircraft face today is that some of the engines fail while the plane accelerates. In that case, the pilot must decide in an instant whether it is better to take flight and solve the problem in the air or if it is preferable not to take off.
Although for many staying on the ground might seem the most sensible option, it is not as simple as it seems: to suddenly decelerate an aircraft, with the weight it has and the speed it reaches can cause accidents. However, today a special cement was designed that runs around the runways of the airports, which when coming into contact with the wheels of the aircraft the ground breaks and helps to slow down.
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
