you throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the gr
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The complete question is;
Within the green dashed circle shown in the figure below, the magnetic field changes with time according to the expression B = 6.00t³ − 1.00t² + 0.800, where B is in teslas, t is in seconds, and R = 2.20 cm.
(a) When t = 2.00 s, calculate the magnitude of the force exerted on an electron located at point P1, which is at a distance r1 = 5.7 cm from the center of the circular field region.
I have attached the figure the question talks about.
Answer:
Force = 4.62 x 10^(-20) N
Explanation:
First of all,
∮E.ds = (d/dt)∮B.dA
E∮ds = (d/dt)B∮dA
dA is the area where B is not equal to zero.
Thus,
E•2πr = (d/dt)B•πR²
|E| = [(d/dt)B•πR²]/2πr
F = |qE| = q[(d/dt)B•πR²]/2πr
Where q is charge on electron = 1.6 x 10^(-19) C
F is magnitude of force exerted on electron
F = 1.6 x 10^(-19)[(d/dt)B•R²]/2r
F = 0.8 x 10^(-19)[(d/dt)B•R²]/r
Now, dB/dt = 18t² - 2t
Thus,
F = 0.8 x 10^(-19)[(18t² - 2t)•R²]/r
Thus, at t=2 and R=2.2cm = 0.022m and r = 5.7cm = 0.057m
Thus,
F = 0.8 x 10^(-19)[(18x2²) - (2x2)•(0.022²)]/(0.057)
F = (0.8 x 10^(-19) x 0.032912)/0.057 = 4.62 x 10^(-20) N
Answer:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Explanation:
Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure (), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:
Where:
,
- Water total pressures inside the tank and at ground level, measured in pascals.
- Water density, measured in kilograms per cubic meter.
- Gravitational acceleration, measured in meters per square second.
,
- Water speeds inside the tank and at the ground level, measured in meters per second.
,
- Heights of the tank and ground level, measured in meters.
Given that ,
,
,
,
and
, the expression is reduced to this:
And final speed is now calculated after clearing it:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
#1)
1). A point on the outside of a spinning top whose rotational speed is constant.
It will not accelerate as its net speed is ZERO
2). A skydiver whose air resistance is equal to that of her weight.
Since weight is counterbalanced by resistance so here net force is zero so acceleration must be zero here
3). A car on the freeway experiencing a net force of -120 N.
Since there it acts net force F = -120 N so it will accelerate here
4). A submerged beach ball whose buoyant force is eight times the force of gravity on it.
Since buoyant force is more than weight of the ball so it will accelerate upwards
#2)
A non accelerating fra me is always a valid inertial frame
A frictionless spinning merry go round.
(since it is having centripetal acceleration so it is non inertial frame)
A falling rock.
(Since it is falling with acceleration a = 9.8 m/s^2 so it is non inertial)
A hot air balloon moving at 30 degrees east of north with no net force.
(Since net force on hot air balloon is ZERO so it is Inertial Frame)
A space shuttle whose boosters just ignited for takeoff.
(Since space shuttle ignite the boosters so it will accelerate here and hence it is non inertial)
#3
here two forces given as
towards right
towards Left
Since left side applied force is more than right side applied force so net force on it will be towards Left and it is given by
So it is 60 N Left