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Korolek [52]
3 years ago
9

In the equation (v2 - v1)

Physics
1 answer:
myrzilka [38]3 years ago
6 0
V1 is the starting speed.
V2 is the ending speed.
'a' is the average acceleration.
It's equal to the change in speed divided by the time for the change.
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Any five chapter 9 institutions found in the constitution of south africa
navik [9.2K]
Five institutions from the chapter 9 found in the constitution of south Africa are : 
- The Public Protector
- The south African Human right commission
- The commission of Gender Equality
- The Auditor- General
- The Independent Electoral Commission

Hope this helps
7 0
3 years ago
Determine the mechanical energy of this object: a 2-kg pendulum has a speed of 1 m/s at a height of 1/2 meter.
Anuta_ua [19.1K]
The answer is B 10.8 j 
3 0
4 years ago
Read 2 more answers
A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce
Karolina [17]

Answer:

F = 352 N

Explanation:

we know that:

F*t = ΔP

so:

F*t = MV_f-MV_i

where F is the force excerted by the wall, t is the time, M the mass of the ball, V_f the final velocity of the ball and V_i the initial velocity.

Replacing values, we get:

F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)

solving for F:

F = 352 N

 

3 0
3 years ago
Trevor places a mass on a spring. When he releases it, the mass and spring move in simple harmonic motion.
777dan777 [17]

The correct answer is D He could pull the mass down farther.

Amplitude, is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. So to increase the amplitude the spring should be pulled down further which can increase the amplitude.

8 0
3 years ago
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Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0
4 years ago
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