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Sonja [21]
4 years ago
5

What is the path of a projectile called? Friction Track Trajectory acceleration

Physics
2 answers:
olga2289 [7]4 years ago
4 0
The answer is trajectory
sergiy2304 [10]4 years ago
3 0
It is trajectory acceleration. A friction track is a device to study motion in low friction environments, I believe. Does this help?
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A parallel-plate capacitor is made from two aluminum-foil sheets, each 5.9 cm wide and 5.6 m long. Between the sheets is a Teflo
Zarrin [17]

Answer:

1.98 x 10⁻⁷ F

Explanation:

w = width of the sheet = 5.9 cm = 0.059 m

L = length of the sheet = 5.6 m

Area of the sheet is given as

A = L w = (5.6) (0.059) = 0.3304 m²

d = distance between the sheets = 3.1 x 10⁻⁵ m

k = dielectric constant of teflon = 2.1

Capacitance is given as

C = \frac{k\epsilon _{o}A}{d}

C = \frac{(2.1)(8.85\times 10^{-12})(0.3304)}{3.1\times 10^{-5}}

C = 1.98 x 10⁻⁷ F

5 0
3 years ago
How long will it take a person walking at 2.1 m/s to travel 13 m?
MrRissso [65]

Answer:

I gonna give you the number so but you need to round 6.19047619048

Explanation:

  • This is a speed formula so you would use the formula speed=distance/time
  • You need to rearrange it to time=distance/speed
  • So you need to divide 13m by 2.1 m/s

7 0
3 years ago
What is the magnitude of the acceleration of an electron at a point where the electric field has magnitude 6377 n/c and is direc
shusha [124]
Use the magnitude acceleration formula .

4 0
3 years ago
during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which f
vovangra [49]

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

3 0
3 years ago
A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res
VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

KE = PE

\frac{1}{2} mv^2 = mgh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(2(55.8*10^{-2}))}

v = 4.67m/s

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

4 0
3 years ago
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