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alexandr402 [8]
2 years ago
6

When a hypothesis is tested many times and supported by data, it becomes a __________. control theory solution conclusion

Physics
1 answer:
Kitty [74]2 years ago
5 0
I think it would then become a theory
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A car moves with a speed 72km/h, the driver uses the brakes, the car stops after 8 seconds, calculate its speed after 10 seconds
vampirchik [111]
The acceleration of the car is solved by subtracting the initial speed from the final speed then dividing the result by the elapsed time.

initial speed = 72 km/hr = 20 m/s

final speed = 0 m/s

elapsed time = 5 seconds

acceleration = (0 m/s – 20 m/s) / 5 s

acceleration = - 20m/s / 5 s

acceleration = -4 m/s^2
8 0
3 years ago
In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
3 years ago
Mechanical energy is the sum of ________ energy and potential energy.apex
crimeas [40]
Mechanical energy is the sum of kinetic energy and potential energy
6 0
3 years ago
Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tas
Nitella [24]

The tension in the string holding the tassel and the vertical will the tension in the string

  • T = 0.1953 N
  • Ф = 34.4 °

<h3>What is the tension in the string holding the tassel. ?</h3>

Generally, the equation for Tension is  mathematically given as

TCos\theta = mg

Therefore

TCos6.58^{o} = 19.8*10^{-3}*9.8

T = 0.1953 N

b).

Where

T* sin \theta = ma

0.1953*Sin6.58 \textdegree  = 19.8*10^{-3}*a

a = 1.13 m/s^2

In conclusion

T* sinФ = ma

2msinФ = ma

2sinФ = a

sin\theta = \frac{a}{2}

\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}

Ф = 34.4 °

In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string

T = 0.1953 N

Ф = 34.4 °

Read more about tension

brainly.com/question/15880959

#SPJ1

4 0
1 year ago
A 0.20 kg mass attached to the end of a spring causes it to stretch 3.0 cm. What is the spring constant? What is the potential e
SVEN [57.7K]

Given that the mass is m = 0.2 kg and the displacement is x = 3 cm = 0.03 m

We have to find the spring constant and potential energy.

The spring constant can be calculated by the formula

\begin{gathered} F=\text{ kx} \\ mg\text{ = kx} \\ k\text{ = }\frac{mg}{x} \end{gathered}

Here, k is the spring constant.

g = 9.8 m/s^2 is the acceleration due to gravity.

Substituting the values, the spring constant will be

\begin{gathered} k=\frac{0.2\times9.8}{0.03} \\ =\text{ 65.33 N/m } \end{gathered}

The potential energy can be calculated as

\begin{gathered} U=\frac{1}{2}kx^2 \\ =\frac{1}{2}\times65.33\text{ }\times(0.03)^2 \\ =\text{ 0.0293 J} \end{gathered}

8 0
1 year ago
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