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andrew-mc [135]
2 years ago
12

Answer the questions to help you understand your parachute and forces experiment. Use the data table below to record your data.

A sample entry has been made for you.
Note: For size, measure the distance between two parallel sides of the hexagon.
Physics
1 answer:
katen-ka-za [31]2 years ago
7 0
Ohh so hexagon is. Tool use the data
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An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see
siniylev [52]

Answer:

3.88 * 10^(-15) J

Explanation:

We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.

First, we get the potential and potential energy:

Electric potential = E * r

E = electric field

r = distance between plates

Potential = 2.2 * 10^6 * 0.011

= 2.42 * 10^4 V

The relationship between electric potential and potential energy is:

P. E. = q*V

q = charge of electron = 1.602 * 10^(-19) C

P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)

P. E. = 3.88 * 10^(-15) J

6 0
3 years ago
An 4-kg ball experiences a force and accelerates at a rate of 1.5 m/s.
ANTONII [103]

Answer:

6.0 N

Explanation:

The strength of a force is expressed as the magnitude of the force in Newton.

The formula to apply here is :

Force= mass * acceleration

F=ma

Mass, m = 4 kg

Acceleration = 1.5 m/s²

Force= 4 *1.5 = 6.0 N

4 0
3 years ago
Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height h ab
Assoli18 [71]

Answer:

EXplained

Explanation:

from conservation of energy

change in potential energy = gain in kinetic energy

so as all he balls are throws from the same height thus the change in potential energy is the same for all the balls thus the gain in kinetic energy is the same for all the balls and as they have the same initial velocity thus the final velocity is the same for all the balls.

6 0
3 years ago
Read 2 more answers
A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-
wariber [46]

Answer:

Part a)

v = 12.45 m/s

Part B)

F_n = 0.05 N

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

mgH = \frac{1}{2}mv^2 + mg(2R)

so the speed at point A is given as

mg(H - 2R) = \frac{1}{2}mv^2

v = \sqrt{2g(H - 2R)}

v = \sqrt{2(9.81)(21.9 - 2\times 7)}

v = 12.45 m/s

Part B)

Now the force equation at point A is given as

F_n + mg = \frac{mv^2}{R}

F_n = \frac{mv^2}{R} - mg[/tex]

F_n = 0.004(\frac{12.45^2}{7} - 9.81)

F_n = 0.05 N

6 0
3 years ago
How much force is required to accelerate 3 kg at 3 m/sec^2
bogdanovich [222]

Answer:

F=?

a=3m/s2

m=3kg

F=ma=[3][3]=9 N

Explanation:

7 0
3 years ago
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