Note that
1 yd = 0.9144 m

Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m

Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) = 9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%

Answer:
The Soccer Field is longer by
8.56 m, or
9.36 yd, or
9.4%

4 0

2 years ago

A weather balloon is rising vertically from a launching pad on the ground. A technician standing 300 feet from the launching pad

avanturin [10]

Answer:

$\dfrac{dh}{dt} =5\ ft/s$

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300 tanθ

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

now $\dfrac{dh}{dt}$ can be written as

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}$

When θ = π/4,

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

$\dfrac{dh}{d\theta } = 600$

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{dh}{dt} =600\times \dfrac{1}{120}$

$\dfrac{dh}{dt} =5\ ft/s$

5 0

2 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$