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asambeis [7]
2 years ago
15

PLEASE HELP!!!!!!!!! 30 POINTS

Physics
1 answer:
ivolga24 [154]2 years ago
4 0

Answer: experiment data is the things you do in the experiment and the result is the answer

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A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the
siniylev [52]

Answer:

K=58.8N/m

Explanation:

From the question we are told that:

Mass M=0.442

Drop distance d=0.150

Generally the equation for Spring Constant is mathematically given by

 K=\frac{2mg}{x}

 K=\frac{2*0.442*9.8}{1.150}

 K=58.8N/m

6 0
3 years ago
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa
Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

5 0
3 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
2 years ago
One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista
Jlenok [28]

Answer:

(a) 0.94 m/s²

(b) g (planet) = 0.096g

Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where  s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

Substituting these values in equation 2

g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

     g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g  is

g (planet) = 0.096g

5 0
3 years ago
A 1/10th scale model of an airplane is tested in a wind tunnel. The reynolds number of the model is the same as that of the full
trasher [3.6K]

Answer:

of the velocity of a full size plane in the air

7 0
3 years ago
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